专题02 完型填空(解析版)
备战2020年中考英语五年真题分类汇编（广东）专题02 完形填空【2019】三、完形填空通该下面短文,掌握其大意,然后在每小题所给的四个选项中,选出一个最佳答案。
并将答题卡对应题目所选的选项涂黑。
Cindy and Anna were best friends. Some days they could spend hours happily together without any argument, but other days they just could not ___16___ on what to do.One day they decided to play in the garden near their school. “Come on, let's play chess,”Anna said.“I don't want to play chess,”Cindy replied.“We always do what you want to do, Cindy. It's my turn to make a ___17___,” Anna said. She wa s getting a little unhappy and ___18___, leaving Cindy alone.Cindy was very angry. ___19___ she got home, she found she still had Anna’s notebook in her schoolbag. “Well, I’m not giving it back to her today. I’m too mad at her, " Cindy thought.The next day at school, their teacher Mr.s Stone ___20___ their notebooks. But Anna didn’t have hers, and she looked ___21___. Cindy knew she should tell Mrs. Stone that she had the notebook, but she was ___22___ mad at Anna.When it was time for lunch, Cindy finally told Mrs. Stone the ___23___. "Thank you for being ___24___, Cindy. I'm sure Anna will be thankful that you have given me her notebook," said Mrs. Stone.Later, Mrs. Stone asked the two girls together and talked with them. Mrs. Stone helped them ___25___ that it was a good idea to take turns to decide the activity. They became best friends again.16. A. agree B. live C. depend D. try【答案】A【解析】句意：但是有时候她们对于接下来做什么不能够达成一致意见。
考查动词辨析和句意。
agree同意，live 居住，depend依靠，try尝试；根据Some days they could spend hours happily together without any argument, but可知是表示转折，前面是不争吵，转折后可知是意见不一致；故选A。
17. A. promise B. project C. decision D. dialogue【答案】C【解析】句意：轮到我做决定了。
考查名词辨析和语境。
promise承诺，project工程，decision决定，dialogue 对话；根据We always do what you want to do, Cindy. It's my turn to________. 可知我们总是做你想做的，轮到我做决定了；故选C。
18. A. went over B. went on C. went by D. went away【答案】D【解析】句意：她有点不高兴走了。
考查动词短语和语境。
went over复习，went on继续，went by走过，went away走开；根据leaving Cindy alone把森碟独自留下，可知是走开了；故选D。
19. A. Because B. After C. Unless D. If【答案】B【解析】句意：她到家后，发现书包里有安娜的笔记本。
考查连词和语境。
Because因为，After在……之后，Unless 除非，If如果，是否；根据she got home和she found she still had Anna’s notebook in her schoolbag.之间的关系，可知是到家后发现笔记本；故选B。
20. A. gave away B. asked for C. handed in D. paid for【答案】B【解析】句意：Stone老师要他们笔记本。
考查动词短语和语境。
gave away捐赠，asked for要求，handed in 上交，paid for支付；根据their teacher Mrs. Stone和their notebooks.可知是老师要他们的笔记本；故选B。
21. A. worried B. normal C. proud D. relaxed【答案】A【解析】句意：她看起来担心。
考查形容词辨析和语境。
worried担心的，normal正常的，proud骄傲的，relaxed放松的；根据Mrs. Stone ___5___ their notebooks. But Anna didn’t have hers,老师要笔记本，但是安娜没有，可知是担心的；故选A。
22. A. still B. never C. usually D. Almost【答案】A【解析】句意：但是她仍然生安娜的气。
考查副词和语境。
still仍然，never从不，usually通常，almost 几乎；根据I’m not giving it back to her today. I’m too mad at her,可知是仍然生她的气；故选A。
23. A. chance B. method C. truth D. rule【答案】C【解析】句意：森碟最后告诉Stone老师事实。
考查名词辨析和语境。
chance机会，method方法，truth事实，rule规则；根据Thank you for being ___9___谢谢你的诚实，可知是告诉老师事实；故选C24. A. patient B. honest C. active D. quiet【答案】B【解析】句意：谢谢你的诚实，森碟。
考查形容词辨析和语境。
patient耐心的，honest诚实的，active积极的，quiet安静的；根据Cindy finally told Mrs. Stone the ___8___告诉老师事实，可知是诚实的；故选B。
25. A. describe B. explain C. guess D. realize【答案】D【解析】句意：Stone老师帮助她们认识到轮流决定做什么活动是一个好主意。
考查动词辨析和语境。
describe 描写，explain解释，guess猜，realize意识到；根据Mrs. Stone helped them 和that it was a good idea to take turns to decide the activity可知是帮助她们意识到轮流决定做什么活动是一个好主意；故选D。
【2018】三、完形填空（本大题有10小题，每小题1分，共10分）通读下面短文，掌握其大意，然后在每小题所给的四个选项中，选出一个最佳答案。
A wealthy man love his son very much. As he wanted his son to lead a happy life, he decided to send him to see a wise old man for his advice on happiness．When the old man learnt about his 46 , he handed the boy an empty bowl and said. “Go to the river miles away and 47 it with water. I will tell you about it 48 no water is spilt（洒）when you reach here.” Although the boy was very surprised at this, he had no choice but to 49 this task．The boy 50 on foot for the river and some time later came back with a bowl of water. The old man asked him, “Did you notice the beautiful flowers along the road and the birds singing the trees?” The boy could say nothing about them because he gave his 51 attention to the bowl in his hands．The old man smiled and said, “Bring me 52 bowl of water, but this time enjoy the flowers and the singing of birds as well.”When he returned, the boy was able to 53 everything he had seen to the old man. But when he looked down at his bowl, he found 54 that most water was gone. He forgot all about his bowl while enjoying the beautiful things along the road．"Well, young man," the old man said. "Enjoy the beauty of the world, but never forget the water in your bowl. This is the 55 of happiness."46．A．research B．promise C．purpose D．experience【答案】C【解析】考查名词．A调查．B承诺．C目的．D经历．句意"老人得知他的__后，递给男孩一个空碗"．根据上一句he decided to send him to see a wise old man for his advice on happiness他决定送他去见一位聪明的老人，向他提幸福的忠告．可知，应该是"目的"．选C．47．A．wash B．fill C．compare D．connect【答案】B【解析】考查动词．A洗．B充满．C比较．D连接．句意"到河边去，把它__水"．根据上一句he handed the boy an empty bowl 他递给男孩一个空碗．可知，应该是"装满"．选B．48．A．if B．until C．unless D．while【答案】A【解析】考查连词．A如果．B直到．C除非．D在…期间．句意"当你到达这里时，__没有水被溅出来，我会告诉你的．"．可知，应该是"如果"．选A．49．A．put out B．pick out C．point out D．carry out【答案】D【解析】考查搭配．A扑灭．B挑选．C指出．D进行；执行；完成．句意"尽管男孩对此感到非常惊讶，但他别无选择，只能__这项任务．"．可知，是"完成"．选D．50．A．set off B．paid off C．got off D．kept off【答案】A【解析】考查搭配．A出发．B付清．C下车．D（使）不接近．句意"那男孩步行___去河边，过了一会儿又带了一碗水回来了．"．可知，应该是"出发"．选A．51．A．public B．weak C．quick D．full【答案】D【解析】考查形容词．A公众的．B虚弱的．C快的．D满的、完整的．句意"那男孩对它们什么也没说，因为他把___注意力放在手中的碗上．"．可知，为了不让碗里的水洒出来，应该是"完整的"注意力．即充分注意．选D．52．A．any B．every C．another D．the other【答案】C【解析】考查形容词．A任何的．B每个．C（三者以上的）另一个．D另一个（通常与one连用）．句意"再给我__一碗水"．这里只另一碗水（不止两碗），用another．选C．53．A．change B．imagine C．examine D．describe【答案】D【解析】考查动词．A改变．B想象．C检查．D描述．句意"当他回来的时候，男孩能把他所见到的一切__给那个老人．"．可知，老人让他享受花和鸟的歌声，应该是"描述"．选D．54．A．lazily B．sadly C．luckily D．excitedly【答案】B【解析】考查副词．A懒惰的．B伤心地．C幸运地．D兴奋地．句意"但当他低头看碗时，他__发现大部分的水都不见了．"．可知，应该是"伤心地"．选B．55．A．result B．cause C．secret D．decision【答案】C【解析】考查名词．A结果．B原因．C秘诀．D决定．句意"这就是幸福的__"．根据第一行he decided to send him to see a wise old man for his advice on happiness．他决定送他去见一位聪明的老人，向他提幸福的忠告．可知，应该是"秘诀"．选C．【2017】三、完形填空（本大题有10小题，每小题10分，共10分）请读下面短文，掌握其大意．然后在每小题所给的四个选项中，选出一个最佳答案，并将答题卡上对应题目所选的选项涂黑．46．（10分）Jack's love for birds started when he made his first birdfeeder about six years ago．He filled the feeder with seeds （种子），put it in his backyard and then （46 ）B started coming．He got really （47）D in birds as more came．Then be joined a local society．As he realized that more and more birds were dying very（48）C，he wanted to try his best to help them．Besides what he has done，he has his own group called Protecting Out Birds．He does （49）D about birds，run a website to teach people about birds and how to help them，and （50）A boxes for birds．Not long ago，he found that the bluebirds were nesting（做窝）in the dead trees which were often cut down，（51）A he began to make bluebird boxes for the birds in order to save them．Now he wants to use these boxes to（52）B the dead trees．He hangs these boxes up in trees and takes（53）D down every week to see if the birds are nesting in them．He also writes articles，hoping that more people will （54）C protecting natural ecosystems（生态系统）．"Researching birds is （55）C to protecting birds，"jack says，"In order to protect birds，we have to learn and really know about the birds．"46．A．bees B．birds C．people D．scientists【答案】B【解析】考查名词词义辨析．A．bees蜜蜂B．birds鸟C．people人D．scientists科学家，根据第一句中写出Jack喜欢鸟，故推测这里应该是小鸟飞来．故选B．47．A．nervous B．weak C．bored D．interested【答案】D【解析】考查形容词词义辨析．A．nervous紧张的B．weak虚弱的C．bored 无聊的D．interested 感兴趣的，根据前文Jack喜欢鸟并主动让小鸟来到他的院子推测他对于小鸟的到来是感兴趣的．故选D．48．A．peacefully B．safely C．quickly D．sadly【答案】C【解析】考查副词词义辨析．A．peacefully平静的B．safely安全的C．quickly快速的D．sadly难过的，根据后面空后Jack想尽力帮助它们，可知越来越多的鸟正在快速死亡．故选C．49．A．practice B．business C．instruction D．research【答案】D【解析】考查名词词义辨析．A．practice练习B．business商业C．instruction说明D．research 研究，根据前后文，Jack建立自己的保护鸟的组织，并教其他人如何保护小鸟，可推测出Jack应该是做了一些研究．故选D．50．A．builds B．breaks C．lifts D．pushes【答案】A【解析】考查动词词义辨析．A 考查动词，A．builds建造B．breaks打破C．lifts举起D．pushes推，根据空后的boxes可知，应该是建造盒子．故选A．51．A．so B．but C．though D．because【答案】A【解析】考查连词词义辨析．A 考查连词，A．so因此B．but但是C．though虽然D．because 因为，根据前后文的意思：不久，他发现蓝鸟在经常被砍伐的死树上做窝，他开始给这些鸟建造蓝鸟盒子来拯救它们．前后为因果关系．故选A．52．A．take good care of B．take the place of C．catch up with D．come up with【答案】B【解析】考查动词短语辨析．A．take good care of好好照顾B．take the place of 代替C．catch up with赶上D．come up with提出；想出，根据前文"他开始给这些鸟建造蓝鸟盒子来拯救它们"可推知此句表示"Jack 想用盒子代替死树"．故选B．53．A．it B．this C．these D．them【答案】D【解析】考查代词词义辨析．A．it它B．this这个C．these这些D．them 他们此空指的是前面的these boxes，应用them代替．故选D．54．A．wait for B．send for C．join in D．hand in【答案】C【解析】考查动词短语辨析．A．wait for等待B．send for派人去请C．join in加入D．hand in 上交wait for等待，send for发送到，join in加入，hand in交上，根据前文Jack为保护鸟所做的事情可推知，此句是Jack希望人们加入进来一起保护生态系统．故选C．55．A．simple B．crazy C．important D．fresh．【答案】C【解析】考查形容词词义辨析．A．simple简单的B．crazy疯狂的C．important重要的D．fresh新鲜的，根据后句"为了保护鸟，我们必须学会并真正了解它们．"可推知此句表示"研究鸟在保护鸟时起到很重要的作用．"故选C．【分析】本文是一篇记叙文．Jack是位爱鸟人士，喜欢鸟的同时也为保护鸟做出了行动．【解答】46．B47．D48．C49．D50．A51．A52．B53．D54．C55．C【2016】三、完形填空（本大题有10小题，每小题10分，共10分）通读下面短文，掌握其大意，然后在每小题所给的四个选项中，选出_个最佳答案，并将答题卡上对应的选项涂黑．46．I'm sure many people are working hard for a lot of money，a big house，a new car，expensive clothes and so on．Those are considered to be symbols of （46）B on the material level．When I was young，I was also （47）C reaching for material success．So I chose a job as a salesman and I did make some pretty good money．But later I realized that was not what I wanted，because I was often forced to sell products that might not be good for customers．I become very unhappy （48）D I could make a lot of money．I no longer felt proud of my job and even （49）A myself for doing things like that．So I （50）B my job and took another job，this time helping people（51）C the poor and the weak．The sense of belonging was great and suddenly I felt successful in my life again，I made less money but I was （52）D with myself．For my own past，success comes from the feeling of satisfaction and joy．I feel successful when I love what I do，not caring so much about （53）C．And I feel the most successful when I （54）B my kindness everywhere go．So，be （55）A to yourself：learn to listen to the voice from the bottom of your heart．And find your own way with a happy heart to achieve your own success．46．A．luck B．success C．safety D．hope【答案】B【解析】考查名词用法．根据上句I'm sure many people are working hard for a lot of money，a big house，a new car，expensive clothes and so on．（我相信很多人都在为很多钱，一个大房子，一辆新车，昂贵的衣服等努力工作）可知这些应该都是成功的物质层面的标志，结合选项所以应该用名词success成功．故选B 47．A．sadly B．hardly C．crazily D．honestly【答案】C【解析】考查副词用法．根据后句So I chose a job as a salesman and I did make some pretty good money．（所以我选择了一份工作作为一个推销员，并挣了很多钱），结合选项的含义，sadly悲伤地；hardly几乎不；crazily疯狂地；honestly诚实地；可知此处应是指作者也在疯狂地达到这种物质上的成功，所以才挣了那么多钱．所以应用副词crazily疯狂地．故选C48．A．if B．since C．until D．although【答案】D【解析】考查连词用法．根据I become very unhappy （我变得很不高兴）和后句I could make a lot of money （我能挣很多钱）可知前后句存在一种看似矛盾的关系，结合选项的含义if如果，表假设；since自从，表时间；until直到，表时间；although虽然，尽管，表让步，引导让步状语从句，结合语境，应该用although，指我变得非常不高兴，虽然我能挣很多钱．故选D49．A．look down upon B．look up to C．look through D．look into【答案】A【解析】考查固定短语用法．根据I no longer felt proud of my job （我不再为我的工作感到骄傲）推出后句应该是指甚至对自己做过的事看不起自己，结合选项应该用固定短语look down upon"轻视，看不起"．故选A50．A．made up B．give up C．set up D．turn up【答案】B【解析】考查固定短语用法．根据and took another job从事另一个工作，可知前面应该是放弃了自己原来的工作，结合选项应该选固定短语give up"放弃"．故选B51．A．in danger B．in order C．in need D．in surprise【答案】C【解析】考查固定短语用法．根据this time helping people（51）the poor and the weak是指这次帮助…的穷人和弱者；结合选项，in danger处于危险中；in order井然有序的；in need在危难中，在危急中；in surprise 惊奇地；可知此处应该用in need才符合语境，指帮助处于危难中的穷人和弱者．故选C52．A．strict B．cheerful C．angry D．satisfied【答案】D【解析】考查形容词用法．根据前句The sense of belonging was great and suddenly I felt successful in my life again（这种归属感是伟大的，突然我觉得在我的生活中成功了）结合后句的转折连词but可知此句是指虽然挣得钱少了，但我对自己感到…结合选项可知此处应该用固定短语be satisfied with对…感到满意；这里指但对自己感到满意．故选D53．A．trust B．love C．money D．health【答案】C【解析】考查名词用法．根据前句I feel successful when I love what I do当我做我喜欢做的事时，我会感到成功，所以可推知后句的意义为不会太在乎钱，所以此处应该用不可数名词money"钱"．故选C 54．A．throw B．spread C．cell D．lend【答案】B【解析】考查动词用法．根据And I feel the most successful when I （54）my kindness everywhere go可知此句要表达的意义为：当我把我的善良…各地时，我感觉到是最成功的时候，所以主语后缺少一个动词；结合选项，throw"扔掉"；spread"传播"；cell名词，电池，细胞；lend借出；结合语境，此处应该用动词spread"传播"，此处是指把善良传播到各地．故选B55．A．true B．brave C．friendly D．terrible．【答案】A【解析】考查形容词用法．根据后句learn to listen to the voice from the bottom of your heart学会倾听你内心的声音，可知前句应是表达要忠于你自己的想法之意，结合选项的含义，true"真实的"；brave"勇敢的"；friendly"友好的"；terrible"糟糕的"；所以此处应该用形容词true，指作真实的自我．故选A【2015】三、完形填空46．Once upon a time，there was a lazy poor living in a small house with spider webs（蜘蛛网）on the walls and mice running around．People （46）B coming into such a dirty place and the poor man was lonely and sad every day．He thought it was poverty（贫困）that（47）A his unhappy life．One day，the poor man dropped in on a wise old man and asked him for （48）C about changing his life．The old man gave him a beautiful vase（花瓶）and said，“This is a magic vase that will bring you （49）B．”The poor man looked at the vase （50）D．Why would he need a vase in his poor house？However，he didn’t want to （51）A such a beautiful vase，so he brought it home on the table．“It．s not right for something so beautiful to be （52）C．”the poor man looked at the vase and thought．Then he picked some wild flowers and put them into it，making it even more beautiful．（53）B he was still not satisfied．“It is not good for such a beautiful thing to stand next to a spider web．”At this，he started to do some cleaning in the house and paint the walls．His house turned into （54）D place immediately．The poor man （55）A．He suddenly realized that in the past it was his laziness that made him poor and unhappy．From then on，he worked hard and his life got better and better．46．A．enjoyed B．avoided C．forgot D．considered【答案】B【解析】考查语境理解及动词辨析．A．enjoyed 喜欢B．avoided 避免C．forgot忘记D．considered把…看做．根据such a dirty place可知，人们避免去这么脏的地方，因此这个人感到孤独和难过．故选B．47．A．led to B．connected to C．made up D．set up【答案】A【解析】考查语境理解及动词短语．A．led to导致B．connected to 连接到C．made up 编造D．set up 建立．根据He thought it was poverty（贫困）可知，他认为是贫穷（贫困），导致来他不幸福的生活．故选：A．48．A．service B．knowledge C．advice D．care【答案】C【解析】考查语境理解及名词辨析．A．service 服务B．knowledge知识C．advice 建议D．care关心，照顾．根据about changing his life可知，这个可怜的人询问一个聪明的老人，问他改变生活的建议．故选C．49．A．peace B．happiness C．mess D．regret【答案】B【解析】考查语境理解及名词．A．peace 和平B．happiness幸福C．mess 混乱D．regret遗憾，根据This is a magic vase 可知，这是一个神奇的花瓶，将带给你幸福．故选B．50．A．sadly B．nervously C．proudly D．surprisedly【答案】D【解析】考查语境理解及副词．A．sadly伤心地B．nervously 紧张地C．proudly骄傲地D．surprisedly 惊奇地根据The poor man looked at the vase 可知，这个穷人惊奇地看着花瓶．故选D．51．A．throw away B．give out C．pay for D．keep off【答案】A【解析】考查语境理解及动词短语．A．throw away 扔掉B．give out 放弃C．pay for付款D．keep off保持根据he didn’t want to 可知，他不想扔掉这个漂亮的花瓶．故选A．52．A．ugly B．Full C．empty D．dirty【答案】C【解析】考查语境理解及形容词．A．ugly 恶心的B．Full满的C．empty空的D．dirty脏的．根据It．s not right for something so beautiful 可知，花瓶空着不好，故选C．53．A．Although B．But C．So D．Because【答案】A【解析】考查语境理解及连词．A．Although虽然，尽管B．But除了C．So 因此D．Because因为．根据he was still not satisfied可知，尽管他仍然不满意．故选A．54．A．large B．dark C．strange D．comfortable【答案】D【解析】考查语境理解及形容词．Alarge 大的B．dark 黑暗的C．strange 陌生的D．comfortable舒服的．根据he started to do some cleaning in the house and paint the walls可知，房子经过打扫，他的房子立刻变成了一个舒适的地方．故选D．55．A．cheered up B．rang up C．sped up D．stayed up．【答案】A【解析】考查语境理解及动词短语．A．cheered up欢呼B．rang up打电话C．sped up加速D．stayed up．停留．根据His house turned into comfortable place immediately．可知，房子变成了一个舒适的地方，因此这个穷人欢呼起来．故选A．。
专题02 语法填空题解题技巧(二)(解析版)(1)
高考英语语法填空考点讲解与真题分析专题02语法填空题解题技巧（二）（二）有给出词的解题思路给出的词主要有动词，名词，形容词和代词，其中动词占多数，其次是名词，再者是形容词，代词偶尔考查。
对这类题要根据该词在句子中的作用确定其形式。
1. 动词给出的词是动词，根据前面对试题特点的分析可从三个方面考虑：填入该动词的某种时态、语态形式；填入该动词的某种非谓语动词形式；填入该动词的名词形式。
1)作谓语（1）确定用谓语动词句子的基本要素是主谓结构，如果句子中没有谓语，则确定要填入谓语动词的某种形式。
（2）确定谓语动词的具体形式谓语动词的形式主要是时态和语态形式，动词的时态、语态要根据时间状语、上下文、语境及句子结构来确定，另外还要注意主谓一致。
1. I __________ (voice) my biggest concern to my mother, “How will I make friends?”（2019北京）【答案】voiced【解析】句意：我向妈妈吐露了我最关心的问题：“我怎样交朋友？”该句前面和后面的句子用的都是一般过去时，所以该句也用一般过去时，故填入过去式voiced。
2. Research on the question _________ (suggest) that，for most students，it doesn't. （2019北京）【答案】suggests/suggested/has suggested【解析】句意：对这一问题的研究表明，对大多数学生来说，答案是否定的。
由语境可知，此处用一般现在时、一般过去时和现在完成时均可。
用一般现在时和现在完成时时要注意主谓一致。
3. Of the nineteen recognized polar bear subpopulations, three are declining, six _________ (be) stable, one is increasing, and nine lack enough data. (2019全国I)【答案】are【解析】句意：在十九个被公认的北极熊亚种群中，三个在退化，六个稳定，一个在增加，九个缺乏足够的数据。
专题02 登山探险-2024届高考英语时文阅读之语法填空专项训练(解析版)
2024届高考英语时文阅读之语法填空专项训练专题02登山探险Why choose the Alps for your next summer阿尔卑斯山夏季探险基础篇adventure?Arctic adventure: skiing in Sweden's most巩固篇北极度假胜地滑雪northerly resortThe best adventure experiences in Lochaber,英国户外之都洛哈伯探险提高篇the outdoor capital of the UK专项微练单句语法填空代词专项训练真题精选高考模拟衔接名校真题演练【原创题】【基础篇】When picturing the Alps, most travelers conjure snowyscenes. Europe’s most extensive mountain range has becomedefined by its world-class ski resorts, regular host to theWinter Olympics. But until the mid 1800s, the Alps were1 summer playground. Immortalized by artistssuch as Turner and Sargent, and beloved of Grand Tourtravelers and a golden era of summit-vanquishingmountaineers, the Alps were where summer was at. During the calendar’s colder months, British travelers were heading to Europe’s southern coastlines, 2 (seek) water cures and Cote d’Azur winter sunshine.Then seasonal habits shifted. Monied Victorian travelers were tempted to winter in the Alps’ cosy, well-catered chalets, 3 (famous) led by enterprising St Moritz hotelier Johannes Badrutt, 4 offered tohost the first have-a-go snow season tourists from England. But it was the advent of affordable package tourism 5 the mid 1900s that sealed the deal for summer supremacy in Europe’s beach resorts. But is the direction of travel once again on the move? “More and more, clients are seeking out Alpine holidays during the summer months,” says Carolyn Addison, head of product at luxury travel company Black Tomato. “People are swerving the crowds in peak winter season and 6 (look) at summertime in the Alps to be immersed in the mountain air, with plenty of ways to be active in nature. We’re seeing a surge in demand for slow travel, relishing the outdoors and escaping the increasingly baking summer heat in other European destinations.”The company says that rooms booked in the Dolomites almost tripled from 2021 to 2022. And while 2021 bookings were a mix of summer and winter stays, for 2022, almost all were for summer. The Alps — like many of Europe’s wilderness areas and national parks — 7 (benefit) from a pandemic-led boom.“We’ve seen an increase in the8 (popular) of summer travel to the mountains over the last few years,” says Joanna Laforge, co-owner of regional specialist tour operator Ski France. “It increased dramatically after Covid-19, with French clientele looking for holiday options when travel 9 (restrict), meaning the idea of open space, fresh air and a change of pace in the mountains proved very popular. This increase is now happening with the UK market, too. 10 (we) chalet and apartments bookings for summer 2023 are already up on last year.”【答案】1. a2. seeking3. famously4. who5. in6. looking7. benefited8. popularity9. was restricted 10. Our【巩固篇】The shadowy figures ambling along the roof of thecovered passageway above the railway line look like fugitivesas they cross the border. But the skis and snowboards they’recarrying tell 1 different story.The ‘renegades’ are, in fact, skiers and boarders2 have just completed the ‘Norvege Svången’(Norwegian Bow): an off-piste descent from the summit ofRiksgränsen ski resort in Sweden that swings 3 (brief) across the international border into neighbouring Norway before returning into its home country, ending at a railway line.I’m observing all this 4 enjoying a beer on the deck of Niehku Mountain Villa, my home for three nights in Riksgränsen, Sweden’s most northerly ski resort, 125 miles north of the Arctic Circle. The award-winning lodge 5 (build) within the curtilage of a former railway shed once used by steam trains on the Ofoten Line. As well as bringing skiers to Arctic Sweden 6 over 100 years, the line has transported iron ore from the world’s biggest iron ore mine in Kiruna to the permanently ice-free port of Narvik in Norway.More than one billion tonnes of iron ore7 (shuttle) along the line since its construction in 1902, and without it Riksgränsen wouldn’t exist, as the town was developed initially to service the railway. Indeed, before it developed as a skiing and outdoor destination, Riksgränsen was briefly abandoned as the residents struggled8 (cope) with the phenomenal snowfalls it receives most winters.9 with the installation of the first ski lifts on the slopes above the railway line in the 1950s, locals and the soon-to-follow visitors were finally able to take advantage of all that snow — as does Niehku Mountain Villa and10 (it) unique heli-ski operation.【答案】1. a2. who3. briefly4. while5. was built6. for7. have been shuttled8. to cope9. But 10. its【提高篇】If you’re looking for adventure in a land of vast, openwaters, sweeping skies and colossal mountains, then Lochabershould be on your radar. With 1 impressive rangeof outdoor activities — be they high-octane pursuits, slowreconnections with nature or fun 2 (experience) forthe whole family — this clean, green Scottish region nestledin the West of the Scottish Highlands is 3 (worth) of its moniker as the Outdoor Capital of the UK.The historic waterfront town of Fort William is the beating heart of Lochaber, acting 4 the centralcompass point for some of the area's most enticing destinations, from the 80-mile-long Great Glen and towering Ben Nevis to the raw, 5 (rug) peninsulas of Morvern and Ardnamurchan. Amidst such dramatic scenery, Lochaber's countless outdoor activities offer the perfect way 6 (explore) a region with adventure at its core.Scotland is the most 7 (mountain) country in the UK, and its towering peak of Ben Nevis is the highest in Britain, at 4,413ft. 8 more technical ascents and descents of the mountain are possible, the classic ‘Pony Track’ route, which zigzags up the main bulk to the summit and9 (it) unbeatable views, is the most popular with hikers here. Always 10 (treat) Ben Nevis seriously and prepare well, as hiking here is no simple weekend wander. Those in doubt should hire a guide — local outfit Abacus Mountain Guides have years of experience helping people achieve their dream of hiking into the heavens.【答案】1. an2. experiences3. worthy4. as5. rugged6.to explore7. mountainous8. Although/Though9. its 10. treat【专项微练：代词】1．（2023春·福建三明·高一三明一中校考期中）Celebrating Naadam with my friend was totally worth ________. (用适当的词填空)【答案】it【详解】考查固定短语。
专题02 代词、介词和介词短语词(解析版)-近三年高考英语真题(2022-2024)分类汇编
代词、介词和介词短语专题02考点01代词1.（2024年浙江卷1月·语法填空）Who knows,perhaps some of the more forward-looking________(one)mayyet come out with a whole range of“just for you”pack sizes with special offers as well.【答案】ones【解析】考查代词。
句意：谁知道呢，也许一些更有远见的人可能还会推出一系列“只为你”的包装尺寸，并提供特别优惠。
代词one意为“一个人”，在some of后应用复数形式。
故填ones。
2.（2024年全国甲卷语法填空）This area,with_______(it)unique and breathtaking natural beauty,must be well preserved for all people of the nation to enjoy-as a national park.【答案】its【解析】考查代词。
句意：这个地方，以其独特而令人惊叹的自然美景，必须被妥善保存供全国人民欣赏。
这里“它们”用形容词性物主代词做定语修饰名词beauty。
3.（2023年全国甲卷改错）In that class,Miss Zhao,our biology teacher,showed we insects on stamps.【答案】we→us【解析】考查代词。
句意：在那节课上，我们的生物老师赵老师给我们看了邮票上的昆虫。
作动词show的宾语，应用宾格us。
故we改为us。
4.（2023年全国乙卷改错）Last Friday my mom decided to color his hair.She studied with all the hair products at the drugstore.【答案】his→her【解析】考查代词。
专题02 完形填空(解析版)---5年(2017-2021)中考1年模拟英语试题分项详解(天津专用)
5年（2017-2021）中考1年模拟英语试题分项详解（天津专用）专题02 完形填空(解析版)一、2021年二、完型填空Think of all the ways that you use your arms and hands. You use 16 to open doors, carry boxes, climb trees and ride bikes.Jessica Cox was born 17 arms. But she didn’t let that 18 her from doing things. She 19 to feed herself, paint and play the piano by using her feet.When she was at 20 , Jessica watched the other students on the playground. She did not have hands to catch balls with, 21 she did not have arms to climb with. Jessica imagined herself as a girl of unusually (不寻常地) great ability. She would 22 over the playground and take her friends into the sky.Many years later, when Jessica 23 she did fly. She learnt to fly a plane! It was 24 work, but Jessica was patient, confident and brave. She controlled the plane with her feet. She made her 25 come true.16．A．that B．them C．her D．it17．A．without B．along C．including D．towards18．A．divide B．add C．lend D．stop19．A．paid B．refused C．learnt D．forgot20．A．school B．bed C．hospital D．town21．A．so B．but C．and D．though22．A．fly B．knock C．fall D．push23．A．set up B．grew up C．gave up D．warmed up 24．A．comfortable B．lazy C．small D．hard25．A．mark B．advice C．dream D．report【答案】16．B 17．A 18．D 19．C 20．A 21．C 22．A 23．B 24．D 25．C【解析】16．句意：你可以用它们开门、搬箱子、爬树和骑自行车。
专题02：说明文(二)-备考2021年高考英语阅读理解体裁分类专练(含解析)
备战2021年高考英语篇章体裁分类专项训练专题02 说明文（二）话题：一、完形填空People are always asking what the most important element to a healthy relationship is. The 1 is there are many. But there is one really important thing that all great and healthy relationships have in common— 2 . Yes, of course, there are some little 3 lies even in the best of relationships— 4 surprises or unexpected birthday parties—but the truth is that happy couples communicate honestly and don’t 5 to protect their own interests. So how can you have more open 6 in your own relationship?First, remember that honesty is the best policy even 7 your partner might not like what you have to say. Yes, it may be 8 to tell your guy that you don’t like one of his friends. But lying or holding things 9 will make things worse in the long run.The next thing to be 10 is to say what you have to say 11 . If your partner feels 12 ,he will be less13 to what you have to say and may even become defensive. Your delivery is of utmost importance when you are discussing sensitive issues. Think through what you want to say 14 before youbring it up. You don’t want to 15 the person that you love. You just want them to be 16 of the truth.One last thing to keep in mind about communicating openly is that it enhances your relationship. If you aren’t honest about any dirty little 17 you may have, they will probably come up eventually and 18 you in the backside. And if you keep things to yourself, your relationship will not be based on the truth, which19 a solid foundation. Hard20 the truth may be sometimes, honesty really is the best policy.1．A．phenomenon B．truth C．theory D．evidence2．A．honesty B．enthusiasm C．cooperation D．persistence3．A．red B．purple C．white D．black4．A．security B．privilege C．anniversary D．ambition5．A．lie B．bargain C．complain D．interact6．A．appreciation B．reputation C．selection D．communication7．A．why B．when C．how D．where8．A．tough B．immediate C．complicated D．plain9．A．in B．on C．above D．back10．A．serious about B．proud of C．expert at D．worthy of11．A．obviously B．rudely C．nicely D．smoothly12．A．puzzled B．attacked C．astonished D．satisfied13．A．creative B．relative C．sensitive D．receptive14．A．slightly B．thoroughly C．frequently D．briefly15．A．suspect B．convince C．insult D．frighten16．A．afraid B．aware C．ashamed D．capable17．A．uniforms B．secrets C．blankets D．courts18．A．train B．consult C．understand D．bite19．A．picks out B．turns on C．makes for D．comes across20．A．although B．while C．until D．as二、阅读选择AArbeia Roman Fort (城堡) and MuseumLocation and HistoryArbeia Roman Fort is situated on Hadrian’s Wall. It was the most important structure built by the Romans in Britain, and now it has been a World Heritage (遗产) Site. Built around AD 160, Arbeia Roman Fort was the military supply base for the soldiers who were stationed along Hadrian’s Wall. The fort has been gradually uncovered and some original parts have been revealed. There are reconstructions that show how Arbeia Roman Fort would have looked.The ReconstructionsThe reconstructions of the Commanding Officer’s house and soldiers’ quarters are strikingly different. The accommodation for soldiers is dark and uncomfortable, while the Commanding Officer’s house is spacious and luxurious, with courtyards with fountains for him and his family to enjoy.The MuseumVisit the museum and see many objects that were found at Arbeia. They are historically important and show what daily life was really like at that time. You will see weapons, tools, jewellery, and so on. You can also discover how the Romans buried their dead and see tombstones (墓碑) which survive to this day. There is a “hands-on” area allowing visitors to dig on a certain site and study their findings with the help of museum staff. You can piece together pottery (陶器), or try writing just as the Romans would have done. For children, they can build this ancient Roman fort with building blocks by themselves.Opening Times and Getting ThereApril 1-October 31:Monday to Saturday 10:00am-5:00pm, Sunday 2:00pm-5:00pm.November 1-March 31:Monday to Saturday 11:00am-4:00pm, closed Sunday. (Closed December 25-26 and January 1)Entry is free.Arbeia is only a ten-minute walk from the bus station at South Shields. Free car park nearby.Website:/arbeia21．According to the passage, Arbeia Roman Fort ______.A. was related to the militaryB. got reconstructions around AD 160C. was built in a small area in RomeD. provided a comfortable life for soldiers22．What can visitors do in the museum?A. See historical objects.B. Build tombstones for the dead.C. Write to ancient Romans.D. Try using old tools and weapons.23. What time does Arbeia Roman Fort close?A. On April 1.B. On October 31.C. On November 1.D. On December 26.BYou know that old saying that laughter is the best medicine. Well, studies have long shown that laughter can have a positive effect both physically and emotionally. In South Korea, a nation more used to holding back its emotions, at least one hospital is encouraging patients to let loose on their regular basis.Laughing, for those cancer patients and their families, is a weekly exercise at Seoul National University hospital. It is something that does not come easy for them, but an hour of laughter therapy (疗法）is all it takes to fight depression that often follows chemical treatment. Lim Song Li, a therapist at the hospital, was once a depression patient herself. She now is a laughter therapist and says when you laugh, blood vessels expand,and sugar levels drop, producing an abundance of hormones (激素）linked with happiness and pleasure. But in Korean culture,where Confucian tradition dominates(主导）social behavior,laughing is not a nature thing. Korean men are taught not to cry more than three times in their lifetime. And the sound of a Korean woman's laughter should not be heard outside the fence of her home. But inside this hospital,they are letting it out. By the end of the session, they make belief laughs somehow become their own.If laughing requires effort,more natural to Koreans, it is singing. The sing-song star therapist, famous for her therapy sessions to fight housewife depression,Jeong Ji Song says singing is an easier way to express inner feelings,especially for Korean women brought up in the conservative background. For some,these classes can be a stress-management tool,but for many more who suffer from depression, learning to sing out their heart can be a healing process.It not only helped Ying Seung Woo come out of severe depression, but also presented her with a new career. After taking up singing therapy, she found a talent in herself-cheer-leading. And now she is taking courses to become a certified therapist.24.The author mentions South Korea to show.A．there are few people suffering from depression in South KoreaB．laughter has a positive influence on Korean women' s healthC．people in South Korea are used to expressing their inner feelingsD．people in South Korea have attempted to use laughter therapy25. Why does the hospital encourage patients to laugh?A. Laughter can help ease patients' pain.ughing can help patients fight depression.C. Laughter is a natural thing in the Korean society.D. Laughing is not allowed in the Korean tradition.26. What does the underlined word “healing” in paragraph 3 mean?A．Curing. B．Controlling. C．Suffering. D．Fighting.CAs doctors and nurses struggle for medical supplies to fight coronavirus pandemic,help is coming from an unlikely place﹣high school students. The Careen Technical Education Charter High Scholl（CTEC）,US,has been using its nine 3D printers to make face masks for healthcare workers on the frontline.Face masks have been particularly challenging to find across the US since the outbreak Valerie Castro,16, a CTEC student,is part of a team of teachers and students who came up with improvement that cut almost 30 minutes off the time it takes to build a face mask."That's what makes this school kind of different. " Gavin Newsom, Governor of the State of California, said. "Even though all these bad things are happening，we're able to make an impact positively. It's like leaving your little mark on the world. " Makerbot，a New York﹣based company that makes 3D printers, has also helped CTEC to maximize the building process. They've been churning out nearly 100 masks each day, and that production rate is set to be three times as 20 more 3D printers come online at the school.CTEC staff and students have provided masks to hospitals, dentists, urgent care centers and retirement homes in Fresno County and the surrounding areas in California. "I was shocked," Stacy V ohra told school staff in video. "We were so thrilled to have the donation from CTEC. This is something that we've been needing. ""The entire staff has switched from daily teaching to face mask production，" said Jonathan Delano, director of CTEC. "When the new 3D printers arrive, some of them will go to students' houses so kids can help with the hands ﹣on process. ""That's how we get through these things, " Delano said. "Our high school focuses on giving back to society. Students should know the skills they hold can have an impact on a community. "27．What can we learn from the passage?A. High School students are expected to help make face masks in the U. S.B. Valeria alone helps speed up the production of 3D face masks in CTEC.C. The act of the students in CTEC is highly thought of by Gavin.D. Makerbot helps CTEC to build a factory for making 3D face masks.28．What does the underlined phrase "churning out" in Paragraph 3 mean?A. producing massiveB. selling quicklyC. importing legallyD. exchanging equally29．How do people feel to receive the donation from CTEC?A. Shocked and embarrassed.B. Astonished and excited.C. Disappointed and frustrated.D. Inspired and joyful.30．What's the main idea of this passage?A. Teenagers helped design face masks.B. 3D printing technology was used to produce face masks.C．A high school made donations to fight the pandemic.D. A high school made contributions by producing 3D face masks.三、七选五If you’re tired of freestyle walking programs, you may be interested in leaming how to racewalk. Or you want to pick up speed. Or maybe you have a desire for competition. 31．, racewalking is a good choice for you.As you begin to racewalk, it is important to concentrate more on movement than on speed. Speed will come later, as you master the movement skills of racewalking. And you can begin to increase your speed once you feel comfortable with the actions of racewalking. Be sure, however, to increase your speed gradually. 32．.Racewalking can increase your walking exercise. You can experience this in person. Try walking as fast as you can, and youll feel your body eager to jog. 33．, you have to keep one foot on the ground at all times. While you are jogging, both feet are off the ground at some point, which allows you to cover more distance with one step. So in order to cover the same distance, you have to take more steps than you would if you were jogging.34．. Before you begin exercising, you need to warm up your body and relax your muscles (肌肉). 35．. It is necessary to cool down in the end. If you ignore these steps, you may suffer muscle pain and even injury.A．No matter what the reason isB．In order to walk at high speedsC．The same is true after each exerciseD．Always keep safety in mind when racewalkingE.In order to continue walking and not to break into a jogF.Don’t push too hard or walk to the point of being tired outG.It is more efficient to warm up your body before exercise四、用单词的适当形式完成短文阅读下面短文，在空白处填入1个单词或括号内单词的正确形式。
部编数学八年级上册专题02模型方法课之截长补短解题方法专练(解析版)(人教版)含答案
专题02模型方法课之截长补短解题方法专练（解析版）学校:___________姓名：___________班级：___________考号：___________一、填空题1．如图，△ABC 中，E 在BC 上，D 在BA 上，过E 作EF ⊥AB 于F ，∠B ＝∠1+∠2，AB ＝CD ，BF ＝43，则AD 的长为________．【答案】83【分析】在FA 上取一点T ，使得FT=BF ，连接ET ，在CB 上取一点K ，使得CK=ET ，连接DK ．想办法证明AT=DK ，DK=BD ，推出BD=AT ，推出BT=AD 即可解决问题．【详解】在FA 上取一点T ，使得FT =BF ，连接ET ，在CB 上取一点K ，使得CK =ET ，连接DK ．∵EB =ET ，∴∠B =∠ETB ，∵∠ETB =∠1+∠AET ，∠B =∠1+∠2，∴∠AET =∠2，∵AE =CD ，ET =CK ，∴△AET ≌△DCK (SAS )，∴DK =AT ，∠ATE =∠DKC ，∴∠ETB =∠DKB ，∴∠B =∠DKB ，∴DB =DK ，∴BD =AT ，∴AD =BT ，∵BT =2BF =83，∴AD =83，故答案为：83．【点睛】本题考查全等三角形的判定和性质，等腰三角形的判定和性质等知识点，解题关键在于学会添加常用辅助线，构造出全等三角形．二、解答题2．如图，ABC D 中，BE ，CD 分别平分ABC Ð和ACB Ð，BE ，CD 相交于点F ，60A Ð=°．（1）求BFD Ð的度数；（2）判断BC ，BD ，CE 之间的等量关系，并证明你的结论．【答案】（1）∠BFD ＝60°；（2）BC ＝BD ＋CE ；证明见解析【分析】（1）根据角平分线和外角性质求解即可；（2）在BC 上截取BG ＝BD ，连接FG ，证明△BDF ≌△BGF ，△CGF ≌△CEF ，即可得到结果；【详解】（1）∵BE ，CD 分别平分ABC Ð和ACB Ð，BE ，∴ABE CBE Ð=Ð，ACD BCD Ð=Ð，∴120ABC ACB Ð+Ð=°，∴60FB C FC B Ð+Ð=°，∴60DFB Ð=°．（2）BC ＝BD ＋CE ；证明方法：在BC 上截取BG ＝BD ，连接FG ，在△BDF 和△BGF 中，BD BG DBF GBF BF BF =ìïÐ=Ðíï=î，∴()△△B DF BG F S A S @，∴60D FB B FG Ð=Ð=°，又∵G C F E C F Ð=Ð，∴△CGF ≌△CEF （ASA ）,∴CE ＝CG ，∴BC ＝BD ＋CE ．【点睛】本题主要考查了三角形内角和定理、外角定理、三角形全等应用，准确分析是解题的关键．3．已知等边三角形ABC ，D 为△ABC 外一点，BDC 120Ð=°，BD=DC ，MDN 60Ð=°，射线DM 与直线AB 相交于点M ，射线DN 与直线AC 相交于点N ．（1）当点M 、N 在边AB 、AC 上，且DM=DN 时，直接写出BM 、NC 、MN 之间的数量关系；（2）当点M 、N 在边AB 、AC 上，且DM ¹DN 时，猜想①中的结论还成立吗？若成立，请证明；（3）当点M 、N 在边AB 、CA 的延长线上时，请画出图形，并求出BM 、NC 、MN【答案】（1）BM+NC=MN ，证明见解析；（2）成立，证明见解析；（3）NC-BM=MN ，证明见解析．【分析】（1）由DM=DN ，∠MDN=60°，可证得△MDN 是等边三角形，又由△ABC 是等边三角形，CD=BD ，易证得Rt △BDM ≌Rt △CDN ，然后由直角三角形的性质，即可求得BM 、NC 、MN 之间的数量关系 BM+NC=MN ；（2）在CN 的延长线上截取CM 1=BM ，连接DM 1．可证△DBM ≌△DCM 1，即可得DM=DM 1，易证得∠CDN=∠MDN=60°，则可证得△MDN ≌△M 1DN ，然后由全等三角形的性质，即可得结论仍然成立；（3）首先在CN 上截取CM 1=BM ，连接DM 1，可证△DBM ≌△DCM 1，即可得DM=DM 1，然后证得∠CDN=∠MDN=60°，易证得△MDN ≌△M 1DN ，则可得NC-BM=MN ．【详解】解（1）BM 、NC 、MN 之间的数量关系：BM+NC=MN ．证明如下：∵BD=DC ，DM=DN ，MDN 60Ð=°∴∠BDC=∠DCB=180302BDC °-Ð=°，△MDN 为等边三角形，∴MN=MD=DN ，∵△ABC 是等边三角形，∴∠ABC=∠ACB=60°，∴∠ABD=∠ACD=90°，∴Rt △BDM ≌Rt △CDN （HL ），∴∠BDM =∠CDN=302BDC MDN Ð-Ð=°，∴11,22BM DM NC DN ==,∴BM+NC=MN ．证明：在CN的反向延长线上截取CM1=BM，连接DM1．∵∠MBD=∠M1CD=90°，BD=CD，∴△DBM≌△DCM1，∴DM=DM1，∠MBD=∠M1CD，∵∠MDN=60°，∠BDC=120°，∴∠M1DN=∠MDN=60°，∴△MDN≌△M1DN，∴MN=M1N=M1C+NC=BM+NC，（3）证明：在CN上截取CM1=BM，连接DM1．与（2）同理可证△DBM≌△DCM1，∴DM=DM1，与（2）同理可证∠CDN=∠MDN=60°，∴△MDN≌△M1DN，∴MN=M1N，∴NC-BM=MN．【点睛】本题考查了等边三角形，直角三角形，等腰三角形的性质以及全等三角形的判定与性质等知识．此题综合性很强，难度较大，解题的关键是注意数形结合思想的应用与辅助线的作法．4．在四边形ABDE中，C是BD边的中点．（1）如图（1），若AC 平分BAE Ð，90ACE Ð=°，则线段AE 、AB 、DE 的长度满足的数量关系为______；（直接写出答案）（2）如图（2），AC 平分BAE Ð，EC 平分AED Ð，若120ACE Ð=°，则线段AB 、BD 、DE 、AE 的长度满足怎样的数量关系？写出结论并证明．【答案】（1）AE ＝AB ＋DE ；（2）AE ＝AB ＋DE ＋12BD ，证明见解析．【分析】（1）在AE 上取一点F ，使AF ＝AB ，由三角形全等的判定可证得△ACB ≌△ACF ，根据全等三角形的性质可得BC ＝FC ，∠ACB ＝∠ACF ，根据三角形全等的判定证得△CEF ≌△CED ，得到EF ＝ED ，再由线段的和差可以得出结论；（2）在AE 上取点F ，使AF ＝AB ，连结CF ，在AE 上取点G ，使EG ＝ED ，连结CG ，根据全等三角形的判定证得△ACB ≌△ACF 和△ECD ≌△ECG ，由全等三角形的性质证得CF ＝CG ，进而证得△CFG 是等边三角形，就有FG ＝CG ＝12BD ，从而可证得结论．【详解】解：（1）如图（1），在AE 上取一点F ，使AF ＝AB ．∵AC 平分∠BAE ，∴∠BAC ＝∠FAC ．在△ACB 和△ACF 中，AB AF BAC FACAC AC ìïÐÐíïî＝＝＝∴BC ＝FC ，∠ACB ＝∠ACF ．∵C 是BD 边的中点，∴BC ＝CD ．∴CF ＝CD ．∵∠ACE ＝90°，∴∠ACB ＋∠DCE ＝90°，∠ACF ＋∠ECF ＝90°．∴∠ECF ＝∠ECD ．在△CEF 和△CED 中，CF CD ECF ECDCE CE ìïÐÐíïî＝＝＝∴△CEF ≌△CED （SAS ）．∴EF ＝ED ．∵AE ＝AF ＋EF ，∴AE ＝AB ＋DE ．故答案为：AE ＝AB ＋DE ；（2）AE ＝AB ＋DE ＋12BD ．证明：如图（2），在AE 上取点F ，使AF ＝AB ，连结CF ，在AE 上取点G ，使EG ＝ED ，连结CG ．∵C 是BD 边的中点，∴CB ＝CD ＝12BD ．∵AC 平分∠BAE ，∴∠BAC ＝∠FAC ．在△ACB 和△ACF 中，AB AF BAC FACìïÐÐ＝＝∴△ACB≌△ACF（SAS）．∴CF＝CB，∠BCA＝∠FCA．同理可证：△ECD≌△ECG∴CD＝CG，∠DCE＝∠GCE．∵CB＝CD，∴CG＝CF．∵∠ACE＝120°，∴∠BCA＋∠DCE＝180°−120°＝60°．∴∠FCA＋∠GCE＝60°．∴∠FCG＝60°．∴△FGC是等边三角形．∴FG＝FC＝12BD．∵AE＝AF＋EG＋FG，∴AE＝AB＋DE＋12BD．【点睛】本题主要考查了全等三角形的判定与性质的运用，能熟练应用三角形全等的判定和性质是解决问题的关键．5．在△ABC中，AB=AC，点D与点E分别在AB、AC边上，DE//BC，且DE=DB，点F与点G分别在BC、AC边上，∠FDG12=∠BDE．（1）如图1，若∠BDE=120°，DF⊥BC，点G与点C重合，BF=1，直接写出BC= ；（2）如图2，当G在线段EC上时，探究线段BF、EG、FG的数量关系，并给予证明；（3）如图3，当G在线段AE上时，直接写出线段BF、EG、FG的数量关系：_____________．【答案】（1）4；（2）FG=BF+EG，见解析；（3）FG=BF-EG【分析】（1）解直角三角形分别求出DF，CF即可解决问题．（2）如图2中，结论：FG=BF+EG．在EA上截取EH，使得EH=BF．利用两次全等，证明FG=GH即可解决问题．（3）如图3中，结论：FG=BF-EG．在射线EA上截取EH，使得EH=BF．利用两次全等，证明FG=GH即可解决问题．【详解】（1）∵DE∥BC，∴∠BDE+∠ABC=180°，∵∠BDE=120°，∴∠ABC=60°，∵DF⊥BF，∴∠BFD=90°，∴DF=BF•tan60°1=，∵∠CDF12=∠BDE=60°，∠DFC=90°，∴CF=DF•tan60°3==，∴BC=BF+CF=1+3=4；（2）如图2中，结论：FG=BF+EG．理由：在EA上截取EH，使得EH=BF．∵AB=AC，∴∠B=∠C，∴∠ADE=∠B ，∠AED=∠C ，∴∠ADE=∠AED ，∴∠DEH=∠B ，在△DBF 和△DEH 中，BF EH B DEH BD DE =ìïÐ=Ðíï=î，∴△DBF ≌△DEH （SAS ），∴DF=DH ，∠BDF=∠EDH ，∵∠FDG 12=∠BDE ，∴∠BDF+∠EDG=∠EDH+∠EDG=∠GDH 12=∠BDE ，∴∠GDF=∠GDH ，在△DGF 和△DGH 中，DF DH GDF GDH DG DG =ìïÐ=Ðíï=î，∴△DGF ≌△DGH （SAS ），∴FG=HG ，∵HG=EG+HE=EG+BF ，∴FG=BF+EG ；（3）如图3中，结论：FG=BF-EG ．理由：在射线EA 上截取EH ，使得EH=BF ．∵AB=AC ，∵DE ∥BC ，∴∠ADE=∠B ，∠AED=∠C ，∴∠ADE=∠AED ，∴∠DEH=∠B ，在△DBF 和△DEH 中，BF EH B DEH BD DE =ìïÐ=Ðíï=î，∴△DBF ≌△DEH （SAS ），∴DF=DH ，∠BDF=∠EDH ，∴∠BDE=∠FDH ，∵∠FDG 12=∠BDE 12=∠FDH ， ∴∠GDF=∠GDH ，在△DGF 和△DGH 中，DF DH GDF GDH DG DG =ìïÐ=Ðíï=î，∴△DGF ≌△DGH （SAS ），∴FG=HG ，∵HG=HE-GE=BF-EG ，∴FG=BF=-EG ．【点睛】本题考查了等腰三角形的性质，全等三角形的判定和性质，解直角三角形等知识，解题的关键是学会添加常用辅助线，构造全等三角形解决问题．6．通过类比联想、引申拓展典型题目，可达到解一题知一类的目的．下面是一个案例，请补充完整．（解决问题）如图，点E 、F 分别在正方形ABCD 的边BC 、CD 上，45EAF Ð=°，连接EF ，则EF BE DF =+，试说明理由．证明：延长CD 到G ，使DG BE =，在ABE △与ADG V 中，90AB AD B ADG BE DG =ìïÐ=Ð=°íï=î∴ABE ADG V V ≌理由：（SAS ）进而证出：AFE △≌___________，理由：（__________）进而得EF BE DF =+．（变式探究）如图，四边形ABCD 中，AB AD =，90BAD Ð=°点E 、F 分别在边BC 、CD 上，45EAF Ð=°．若B Ð、D Ð都不是直角，则当B Ð与D Ð满足等量关系________________时，仍有EF BE DF =+．请证明你的猜想．（拓展延伸）如图，若AB AD =，90¹°∠BAD ，45EAF й°，但12EAF BAD Ð=Ð，90B D Ð=Ð=°，连接EF ，请直接写出EF 、BE 、DF之间的数量关系．【答案】（1）AFE AFG △≌△，理由：SAS ；（2）180B D Ð+Ð=°，证明见解析；（3）BE+DF=EF ．【分析】（1）在前面已证的基础上，得出结论AE AG =，进而证明AFE AFG △≌△，从而得出结论；（2）利用“解决问题”中的思路，同样去构造AFE AFG △≌△即可；（3）利用前面两步的思路，证明全等得出结论即可．【详解】（1）ABE ADG Q V V ≌，,,AE AG BAE DAG BE DG \=Ð=Ð=，则BAE FAD FAD ADG FAG Ð+Ð=Ð+Ð=Ð，45EAF Ð=°Q ，45FAG \Ð=°，在AFG V 与AFE △中，AE AG EAF GAFAF AF =ìï=íï=î∠∠AFE AFG \△≌△，理由：（SAS ）EF FG FD DG FD BE \==+=+；（2）满足180B D Ð+Ð=°即可，证明如下：如图，延长FD 至G ，使BE DG =，180B ADF Ð+Ð=°Q ，180ADF ADG Ð+Ð=°，B ADG \Ð=Ð，在ABE △与ADG V 中，AB AD B ADGBE DG =ìïÐ=Ðíï=î()ABE ADG SAS \V V ≌，,,AE AG BAE DAG BE DG \=Ð=Ð=，则BAE FAD FAD ADG FAG Ð+Ð=Ð+Ð=Ð，45EAF Ð=°Q ，45FAG \Ð=°，在AFG V 与AFE △中，AE AG EAF GAFAF AF =ìï=íï=î∠∠AFE AFG \△≌△，理由：（SAS ）EF FG FD DG FD BE \==+=+；（3）BE+DF=EF ．证明如下：如图，延长FD 至G ，使BE DG =，在ABE △与ADG V 中，90AB AD B ADG BE DG =ìïÐ=Ð=°íï=î()ABE ADG SAS \V V ≌，,AE AG BAE DAG \=Ð=Ð，则BAE FAD FAD ADG FAG Ð+Ð=Ð+Ð=Ð，12EAF BAD Ð=ÐQ ，12FAG EAD FAE \Ð=Ð=Ð，在AFG V 与AFE △中，AE AG EAF GAFAF AF =ìï=íï=î∠∠AFE AFG \△≌△，理由：（SAS ）EF FG FD DG FD BE \==+=+；．【点睛】本题考查了截长补短的方法构造全等三角形，能够理解前面介绍的方法并继续探究是解决问题的关键．7．阅读题：如图1，OM 平分AOB Ð，以O 为圆心任意长为半径画弧，交射线OA ，OB 于C ，D 两点，在射线OM 上任取一点E （点O 除外），连接CE ，DE ，可证OCE ODE △△≌，请你参考这个作全等的方法，解答下列问题：（1）如图2，在ABC V 中，2A B Ð=Ð，CD 平分ACB Ð交AB 于点D ，试判断BC 与AC 、AD 之间的数量关系；（2）如图3，在四边形ABCD 中，AC 平分BAD Ð，10BC CD ==，20AB =，8AD =，求ABC V 的面积．【答案】（1）BC=AC+AD ；（2）△ABC 的面积为80．【分析】（1）在CB 上截取CE=CA ，则由题意可得AD=DE ，∠CED=∠A ，再结合∠A=2∠B 可得DE=BE ，从而得到BC=AD+AC ；（2）在AB 上截取AE=AD ，连结CE ，过C 作CF ⊥AB 于F 点，由题意可得EC=BC ，从而得到EF 的长度，再由勾股定理根据EC 、EF 的长度求得CF 的长度，最后根据面积公式可以得到解答 ．【详解】解：（1）如图，在CB 上截取CE=CA ，则由题意得：△CAD ≌△CED ，∴AD=DE，∠CED=∠A，∵∠A=2∠B，∴∠CED=2∠B，又∠CED=∠B+∠EDB，∴∠B+∠EDB=2∠B，∴∠EDB=∠B，∴DE=BE，∴BC=BE+CE=DE+CE=AD+AC；（2）如图，在AB上截取AE=AD，连结CE，过C作CF⊥AB于F点，∴由题意可得：△CDA≌△CEA，∴EC=CD=BC=10，AE=AD=8，∵CF⊥AB，∴EF=FB=208622AB AE--==，∴8 CF===，∴112088022ABCS AB CF=´=´´=V．【点睛】本题考查三角形全等的综合运用，熟练掌握三角形全等的判定和性质、等腰三角形的判定和性质、勾股定理是解题关键．8．（1）问题背景：如图1：在四边形ABCD中，AB＝AD，∠BAD＝120°，∠B＝∠ADC ＝90°，E、F分别是BC，CD上的点且∠EAF＝60°，探究图中线段BE、EF、FD之间的数量关系．小王同学探究此问题的方法是，延长FD到点G．使DG＝BE．连结AG，先证明VABE ≌V ADG ，再证明V AEF ≌V AGF ，可得出结论，他的结论应是______________；（2）探索延伸：如图2，若在四边形ABCD 中，AB ＝AD ，∠B +∠D ＝180°．E ，F 分别是BC ，CD 上的点，且∠EAF 12=∠BAD ，上述结论是否仍然成立，并说明理由；（3）实际应用：如图3，在某次军事演习中，舰艇甲在指挥中心（O 处）北偏西30°的A 处，舰艇乙在指挥中心南偏东70°的B 处，并且两舰艇到指挥中心的距离相等，接到行动指令后，舰艇甲向正东方向以45海里/小时的速度前进，同时舰艇乙沿北偏东50°的方向以60海里/小时的速度前进，2小时后，指挥中心观测到甲、乙两地分别到达E 、F 处，且两舰艇之间的夹角为70°，试求此时两舰艇之间的距离．【答案】（1）EF ＝BE +DF ；（2）结论EF ＝BE +DF 仍然成立；（3）此时两舰艇之间的距离是210海里【分析】（1）延长FD 到点G ，使DG=BE ．连结AG ，即可证明V ABE ≌V ADG ，可得AE=AG ，再证明V AEF ≌V AGF ，可得EF=FG ，即可解题；（2）延长FD 到点G ，使DG=BE ．连结AG ，即可证明V ABE ≌V ADG ，可得AE=AG ，再证明V AEF ≌V AGF ，可得EF=FG ，即可解题；（3）连接EF ，延长AE 、BF 相交于点C ，然后与（2）同理可证．【详解】解：（1）EF ＝BE +DF ，证明如下：在V ABE 和V ADG 中，DG BE B ADG AB AD =ìïÐ=Ðíï=î，∴V ABE ≌V ADG （SAS ），∴AE ＝AG ，∠BAE ＝∠DAG ，∵∠EAF 12=∠BAD ，∴∠GAF ＝∠DAG +∠DAF ＝∠BAE +∠DAF ＝∠BAD ﹣∠EAF ＝∠EAF，∴∠EAF ＝∠GAF ，在V AEF 和V GAF 中，AE AG EAF GAF AF AF =ìïÐ=Ðíï=î，∴V AEF ≌V AGF （SAS ），∴EF ＝FG ，∵FG ＝DG +DF ＝BE +DF ，∴EF ＝BE +DF ；故答案为 EF ＝BE +DF ．（2）结论EF ＝BE +DF 仍然成立；理由：延长FD 到点G ．使DG ＝BE ．连结AG ，如图2，在V ABE 和V ADG 中，DG BE B ADG AB AD =ìïÐ=Ðíï=î，∴V ABE ≌V ADG （SAS ），∴AE ＝AG ，∠BAE ＝∠DAG ，∵∠EAF 12=∠BAD ，∴∠GAF ＝∠DAG +∠DAF ＝∠BAE +∠DAF ＝∠BAD ﹣∠EAF ＝∠EAF ，∴∠EAF ＝∠GAF ，在V AEF 和V GAF 中，AE AG EAF GAF AF AF =ìïÐ=Ðíï=î，∴V AEF ≌V AGF （SAS ），∴EF ＝FG，∵FG ＝DG +DF ＝BE +DF ，∴EF ＝BE +DF ；（3）如图3，连接EF ，延长AE 、BF 相交于点C ，∵∠AOB ＝30°+90°+（90°﹣70°）＝140°，∠EOF ＝70°，∴∠EOF 12=∠AOB ，又∵OA ＝OB ，∠OAC +∠OBC ＝（90°﹣30°）+（70°+50°）＝180°，∴符合探索延伸中的条件，∴结论EF ＝AE +BF 成立，即EF ＝2×（45+60）＝210（海里）．答：此时两舰艇之间的距离是210海里．【点睛】本题考查了全等三角形的判定以及全等三角形对应边相等的性质，本题中求证△AEF ≌△AGF 是解题的关键．9．在ABC V 中，60ABC Ð=°，点D 、E 分别在AC 、BC 上，连接BD 、DE 和AE ；并且有AB BE =，AED C Ð=Ð．（1）求CDE Ð的度数；（2）求证：AD DE BD +=．【答案】（1）60°；（2）见解析【分析】（1）由AB BE =，60ABC Ð=°，可得ABE △为等边三角形，由AEB EAC C Ð=Ð+Ð，CDE EAC AED Ð=Ð+Ð，AED C Ð=Ð，可证60CDE AEB Ð=Ð=° （2）延长DA 至F ，使AF DE =，连接FB ， 由60BED AED Ð=°+Ð，60BAF C Ð=°+Ð，且C AED Ð=Ð，可证()FBA DBE SAS V V ≌ 由=DB FB ，可证FBD V 为等边三角形，可得BD FD =， 可推出结论，【详解】解：（1）∵AB BE =，60ABC Ð=°，∴ABE △为等边三角形，∴60BAE AEB Ð=Ð=°，∵AEB EAC C Ð=Ð+Ð，CDE EAC AED Ð=Ð+Ð，∵AED C Ð=Ð，∴60CDE AEB Ð=Ð=°（2）如图，延长DA 至F ，使AF DE =，连接FB ， 由（1）得ABE △为等边三角形，∴60AEB ABE Ð=Ð=°，∵60BED AEB AED AED Ð=Ð+Ð=°+Ð，又∵60BAF ABE C C Ð=Ð+Ð=°+Ð，且C AED Ð=Ð，∴BED BAF Ð=Ð，在FBA V 与DBE V 中，AB BE BAF BEDAF DE =ìïÐ=Ðíï=î∴()FBA DBE SAS V V ≌∴=DB FB ，DBE FBAÐ=Ð∴DBE ABD FBA ABD Ð+Ð=Ð+Ð，∴60ABE FBD Ð=Ð=°又∵=DB FB ，∴FBD V 为等边三角形∴BD FD =，又∵FD AF AD =+，且AF DE =，∴FD DE AD BD =+=，【点睛】本题考查等边三角形的判定与性质，三角形全等判定与性质，线段和差，三角形外角性质，关键是引辅助线构造三角形全等证明等边三角形．10．如图，在△ABC 中，AB ＝AC ，∠BAC ＝30°，点D 是△ABC 内一点，DB ＝DC ，∠DCB ＝30°，点E 是BD 延长线上一点，AE ＝AB ．（1）求∠ADB 的度数；（2）线段DE ，AD ，DC 之间有什么数量关系？请说明理由．【答案】（1）120°；（2）DE ＝AD +CD ，理由见解析【分析】（1）根据三角形内角和定理得到∠ABC ＝∠ACB ＝75°，根据全等三角形的性质得到∠BAD ＝∠CAD ＝15°，根据三角形的外角性质计算，得到答案；（2）在线段DE 上截取DM ＝AD ，连接AM ，得到△ADM 是等边三角形，根据△ABD ≌△AEM ，得到BD ＝ME ，结合图形证明结论【详解】解：（1）∵AB ＝AC ，∠BAC ＝30°，∴∠ABC ＝∠ACB ＝12（180°﹣30°）＝75°，∵DB ＝DC ，∠DCB ＝30°，∴∠DBC ＝∠DCB ＝30°，∴∠ABD ＝∠ABC ﹣∠DBC ＝45°，在△ABD 和△ACD 中，AB AC DB DC AD AD =ìï=íï=î，∴△ABD ≌△ACD （SSS ），∴∠BAD ＝∠CAD ＝12∠BAC ＝15°，∴∠ADE ＝∠ABD +∠BAD ＝60°，∴∠ADB ＝180°﹣∠ADE ＝180°﹣60°＝120°；（2）DE ＝AD +CD，理由如下：在线段DE 上截取DM ＝AD ，连接AM ，∵∠ADE ＝60°，DM ＝AD ，∴△ADM 是等边三角形，∴∠ADB ＝∠AME ＝120°．∵AE ＝AB ，∴∠ABD ＝∠E ，在△ABD 和△AEM 中，ABD E ADB AME AB AE Ð=ÐìïÐ=Ðíï=î，∴△ABD ≌△AEM （AAS ），∴BD ＝ME ，∵BD ＝CD ，∴CD ＝ME ．∵DE ＝DM +ME ，∴DE ＝AD +CD ．【点睛】本题考查的是全等三角形的判定和性质、等边三角形的判定和性质，掌握全等三角形的判定定理和性质定理是解题的关键．11．如图，120CAB ABD Ð+Ð=°，AD 、BC 分别平分CAB Ð、ABD Ð，AD 与BC 交于点O ．（1）求AOB Ð的度数；（2）说明AB AC BD =+的理由．【答案】（1）120°；（2）见解析【分析】（1）根据角平分线的定义可得∠OAB+∠OBA=60°，从而得到∠AOB；（2）在AB上截取AE=AC，证明△AOC≌△AOE，得到∠C=∠AEO，再证明∠C+∠D=180°，从而推出∠BEO=∠D，证明△OBE≌△OBD，可得BD=BE，即可证明AC+BD= A B．【详解】解：（1）∵AD，BC分别平分∠CAB和∠ABD，∠CAB+∠ABD=120°，∴∠OAB+∠OBA=60°，∴∠AOB=180°-60°=120°；（2）在AB上截取AE=AC，∵∠CAO=∠EAO，AO=AO，∴△AOC≌△AOE（SAS），∴∠C=∠AEO，∵∠C+∠D=（180°-∠CAB-∠ABC）+（180°-∠ABD-∠BAD）=180°，∴∠AEO+∠D=180°，∵∠AEO+∠BEO=180°，∴∠BEO=∠D，又∠EBO=∠DBO，BO=BO，∴△OBE≌△OBD（AAS），∴BD=BE，又AC=AE，∴AC+BD=AE+BE=A B．【点睛】本题考查了角平分线的定义，三角形内角和，全等三角形的判定和性质，解题的关键是截取AE=AC，利用全等三角形的性质证明结论．12．如图所示，已知△ABC中AB＞AC，AD是∠BAC的平分线，M是AD上任意一点，求证：MB－MC＜AB－AC．【答案】见解析【分析】因为AB ＞AC ，所以在AB 上截取线段AE ＝AC ，则BE ＝AB －AC ，连接EM ，在△BME 中，显然有MB －ME ＜BE ，再证明ME ＝MC ，则结论成立．【详解】证明：在AB 上截取AE ＝AC ，连接ME ，在△MBE 中，MB －ME ＜BE （三角形两边之差小于第三边）,∵AD 是∠BAC 的平分线，∴BAD CAD Ð=Ð,在△AMC 和△AME 中，∵AC AE CAM EAMAM AM =ìïÐ=Ðíï=î∴△AMC ≌△AME （SAS ），∴MC ＝ME （全等三角形的对应边相等）．又∵BE ＝AB －AE ，∴BE ＝AB －AC ，∴MB －MC ＜AB －AC ．【点睛】本题考查全等三角形的判定和性质，三角形三边关系以及截长补短法，解题关键是作辅助线构造全等三角形．13．如图所示，已知AC 平分∠BAD ，180B D Ð+Ð=°，CE AB ^于点E ，判断AB 、AD 与BE 之间有怎样的等量关系，并证明．【答案】2AB AD BE =+，证明见解析【分析】在AB 上截取EF ，使EF=BE ，联结CF ．证明()BCE ECF SAS V V ≌，得到B BFC Ð=Ð，又证明AFC ADC V V ≌，得到AF AD =，最后结论可证了．【详解】证明:在AB 上截取EF ，使EF=BE ，联结CF ．CE AB^Q 90BEC FEC \Ð=Ð=°在BCE V 和ECF △BE EF BEC FECCE CE =ìïÐ=Ðíï=î()BCE ECF SAS \V V ≌B BFC \Ð=Ð 180BD Ð+Ð=°Q 180BFC AFC Ð+Ð=°Q 又D AFC \Ð=ÐQ AC 平分∠BADFAC DAC\Ð=Ð在AFC △ 和ADC V中AFC D FAC DACAC AC Ð=ÐìïÐ=Ðíï=î()AFC ADC AAS \V V ≌AF AD\=AB AF BE EF=++Q 2AB AD BE\=+【点睛】本题考查三角形全等知识的综合应用，关键在于寻找全等的条件，作适当的辅助线加以证明．14．如图所示，//AB DC AB AD BE ^，，平分ABC CE Ð，平分BCD Ð；（1）求AB CD 、与BC 的数里关系，并说明你的理由．（2）若把AB AD ^条件去掉，则（1）中AB CD 、与BC 的数里关系还成立吗?并说明你的理由．【答案】（1）AB CD BC +=，见解析；（2）成立，见解析【分析】（1）先写出数量关系，过E 作EF BC ^于F ，然后证明CDE CFE D @D 和ABE FBE @D D ，便可得结论了．（2）成立, 在BC 上截取CF CD =证明CDE CFE D @D 和ABE FBE @D D ，便可得到结论．【详解】()1AB CD BC+=理由是：过E 作EF BC ^于FCE 为角平分线DCE FCE \Ð=Ð//AB DC AB AD^Q，90D \Ð=oEF BC^Q D CFE \Ð=ÐCE CE =Q()CDE CFE AAS D @D CD CF\=同理可证()ABE FBE AAS D @D AB BF \=CF BF AB+=AB CD BC\+=()2成立理由：在BC 上截取CF CD=CE 为角平分线DCE FCE\Ð=ÐCE CE=Q ()CDE CFE SAS D @D CD CF \= D CFE Ð=ÐQ //AB DC180D A \Ð+Ð=o又180CFE EFB Ð+=o QA EFB \Ð=Ð又BE Q 是角平分线ABE FBE \Ð=ÐBE BE =Q()BAE BFE AAS D @D AB FB\=\ CF BF AB+=AB CD BC\+=15．如图，ABC V 是边长为1的等边三角形，BD CD =，120BDC Ð=°，点E ，F 分别在AB ，AC 上，且60EDF Ð=°，求AEF V 的周长．【答案】2【分析】延长AC 至点P ，使CP BE =，连接PD ，证明()BDE CDP SAS △△推出DE DP =，BDE CDP Ð=Ð，进而得到60EDF PDF Ð=Ð=°，从而证明()DEF DPF SAS ≌△△，推出EF=CP ，由此求出AEF V 的周长=AB+AC 得到答案.【详解】解：如图，延长AC 至点P ，使CP BE =，连接PD ．∵ABC V 是等边三角形，∴60ABC ACB Ð=Ð=°．∵BD CD =，120BDC Ð=°，∴30DBC DCB Ð=Ð=°，∴90EBD DCF Ð=Ð=°，∴90DCP DBE Ð=Ð=°．在BDE V 和CDP V 中，BD CD DBE DCP BE CP =ìïÐ=Ðíï=î，∴()BDE CDP SAS △△，∴DE DP =，BDE CDP Ð=Ð．∵120BDC Ð=°，60EDF Ð=°，∴60BDE CDF Ð+Ð=°，∴60CDP CDF Ð+Ð=°，∴60EDF PDF Ð=Ð=°．在DEF V 和DPF V 中，DE DP EDF PDF DF DF =ìïÐ=Ðíï=î，∴()DEF DPF SAS ≌△△，∴EF FP =，∴EF FC BE =+，∴AEF V 的周长2AE EF AF AB AC =++=+=.【点睛】此题考查全等三角形的判定及性质，等边三角形的性质，等腰三角形等边对等角的性质，题中辅助线的引出是解题的关键.16．已知，90POQ Ð=o ，分别在边OP ，OQ 上取点A ，B ，使OA OB =，过点A 平行于OQ 的直线与过点B 平行于OP 的直线相交于点C ．点E ，F 分别是射线OP ，OQ 上动点，连接CE ，CF ，EF ．（1）求证：OA OB AC BC ===；（2）如图1，当点E ，F 分别在线段AO ，BO 上，且45ECF Ð=o 时，请求出线段EF，AE ，BF 之间的等量关系式；（3）如图2，当点E ，F 分别在AO ，BO 的延长线上，且135ECF Ð=o 时，延长AC 交EF 于点M ，延长BC 交EF 于点N ．请猜想线段EN ，NM ，FM 之间的等量关系，并证明你的结论．【答案】（1）见解析；（2）EF AE BF =+；（3）222MN EN FM =+，见解析【分析】（1）连接AB ,通过90POQ Ð=o ，OA OB =得到AOB V 为等腰直角三角形，进而得到45OAB OBA Ð=Ð=o ，根据过点A 平行于OQ 的直线与过点B 平行于OP 的直线相交于点C ，可推出45CBA Ð=o ，45BAC Ð=o ，最后通过证明AOB V ≌ACB △，可以得出结论；（2）在射线AP 上取点D ，使AD BF =，连接CD ，通过证明CAD V ≌CBF V ，得到CD CF =，ACD BCF Ð=Ð，再结合45ECF Ð=o ，90ACB Ð=o 推导证明ECD V ≌ECF △，得到ED EF =，最后等量代换线段即可求解；（3）延长AO 到点D ，使得AD BF =，连接CD ，通过证明CAD V ≌CBF V ，得到CD CF =，ACD BCF Ð=Ð，再结合135ECF Ð=o ，推导证明ECD V ≌ECF △，得到D CFM Ð=Ð，根据D CFB Ð=Ð，等量代换可知CFM CFB Ð=Ð，又因为//AC OQ ，推出MCF CFB Ð=Ð，进而得到MC MF =，同理可证CN EN =，最后根据勾股定理即可求解．【详解】解：（1）证明：连接AB ．Q 90POQ Ð=o ，OA OB =，\AOB V 为等腰直角三角形，\45OAB OBA Ð=Ð=o ，又Q //BC OP ，且90POQ Ð=o ，\BC OQ ^，\90CBF Ð=o ，\45CBA Ð=o ，同理，45BAC Ð=o ，在AOB V 与ACB △中OAB CAB AB ABOBA CBF Ð=Ðìï=íïÐ=Ðî，\AOB V ≌ACB △()ASA ，\90AOB ACB Ð=Ð=o ，OA OB AC BC ===；（2）如图1，在射线AP 上取点D ，使AD BF =，连接CD ．在CAD V 与CBF V 中CA CB CAD CBF AD BF =ìïÐ=Ðíï=î，\CAD V ≌CBF V ()SAS ，\CD CF =，ACD BCF Ð=Ð，Q 45ECF Ð=o ，90ACB Ð=o ，\45ACE BCF Ð+Ð=o ，\45ACE ACD ECD Ð+Ð=Ð=o ，\ECD ECF Ð=Ð，在ECD V 与ECF △中CD CF ECD ECFCE CE =ìïÐ=Ðíï=î\ECD V ≌ECF △()SAS ，\ED EF =，又Q ED AD AE BF AE =+=+，\EF AE BF =+．（3）222MN EN FM =+．证明如下：如图2，延长AO 到点D ，使得AD BF =，连接CD ．\90CAD CBF Ð=Ð=o ，在CAD V 与CBF V 中CA CB CAD CBF AD BF =ìïÐ=Ðíï=î，\CAD V ≌CBF V ()SAS ，\CD CF =，ACD BCF Ð=Ð，Q 90ACD DCB Ð+Ð=o ，\90BCF DCB DCF Ð+Ð==Ðo ，\90FCD BCA Ð=Ð=o ，Q 135ECF Ð=o ，\36090135135ECD Ð=--=o o o o ，\ECF ECD Ð=Ð，在ECD V 与ECF △中EC EC ECD ECF CD CF =ìïÐ=Ðíï=î，\ECD V ≌ECF △()SAS ，\D CFM Ð=Ð，Q CAD V ≌CBF V ，\D CFB Ð=Ð，\CFM CFB Ð=Ð，Q //AC OQ ，\MCF CFB Ð=Ð，\CFM MCF Ð=Ð，\MC MF =，同理可证：CN EN =，\在Rt MCN △中，由勾股定理得：22222MN CN CM EN FM =+=+．【点睛】本题综合考查了全等三角形的性质和判定,勾股定理以及正方形的有关知识，通过添加辅助线构造全等三角形，通过证明全等三角形得到线段之间的关系是解题的关键．17．本学期，我们学习了三角形相关知识，而四边形的学习，我们一般通过辅助线把四边形转化为三角形，通过三角形的基本性质和全等来解决一些问题．（1）如图1，在四边形ABCD 中，AB AD =，180B D Ð+Ð=°，连接AC ．①小明发现，此时AC 平分BCD Ð．他通过观察、实验，提出以下想法：延长CB 到点E ，使得BE CD =，连接AE ，证明ABE ADC △≌△，从而利用全等和等腰三角形的性质可以证明AC 平分BCD Ð．请你参考小明的想法，写出完整的证明过程．②如图2，当90BAD Ð=°时，请你判断线段AC ，BC ，CD 之间的数量关系，并证明．（2）如图3，等腰CDE △、等腰ABD △的顶点分别为A 、C ，点B 在线段CE 上，且180ABC ADC Ð+Ð=°，请你判断DAE Ð与DBE Ð的数量关系，并证明．【答案】（1）①见解析；②CD BC +=，证明见解析；（2）2DAE DBE Ð=Ð，证明见解析【分析】（1）①参考小明的想法，延长CB 到点E ，使得BE CD =，连接AE ，证明ABE ADC △≌△，从而利用全等和等腰三角形的性质可以证明AC 平分；②沿用①中辅助线，延长CB 到点E ，使得BE CD =，连接AE ，证得直角三角形CAE ，再利用勾股定理可求得AC ，BC ，CD 之间的数量关系；（2）类比（1）中证明的思路，延长CD 至F ，使得DF CB =，连AF ，证明ABC ADF ≌△△、ACD ACE V V ≌，再利用全等三角形的对应角相等和等腰三角形等边对等角的性质，找到DAE Ð与DBE Ð的数量关系．【详解】（1）如图，延长CB 到点E ，使得BE CD =，连接AE ．180ADC ABC Ð+Ð=°Q ，180ABE ABC Ð+Ð=°，ADC ABE\Ð=Ð在ADC V 与ABE △中，AD AB ADC ABECD EB =ìïÐ=Ðíï=îQ ()ADC ABE SAS \△≌△ACD AEB \Ð=Ð，AC AE=ACB AEB\Ð=ÐACD ACB \Ð=Ð．AC \平分BCDÐ（2）CD BC +=证明：如图，延长CB 到点E ，使得BE CD =，连接AE ．由（1）知，(SAS)ADC ABE V V ≌DAC BAE \Ð=Ð，AC AE=90BAD DAC CAB Ð=Ð+Ð=°Q90CAE BAE CAB DAC CAB BAD \Ð=Ð+Ð=Ð+Ð=Ð=°在直角三角形CAE 中，90CAE Ð=°2CE \==CD BC \+=（3）2DAE DBEÐ=Ð证明：如图，延长CD 至F ，使得DF CB =，连AF ，由（1）知，()ABC ADF SAS △≌△AF AC \=，ACB FÐ=ÐACD F\Ð=ÐACD ACE\Ð=Ð在ACD △与ACE V 中，CD CEACD ACEAC AC=ìïÐ=Ðíï=îQ ()ACD ACE SAS \△≌△AD AE\=AD AE AB\==ADB ABD \Ð=Ð，AEB ABEÐ=Ð1802BAD ADB \Ð=°-Ð，1802BAE ABE Ð=°-Ð，360DAE BAD BAEÐ=°-Ð-ÐQ ()()36018021802DAE ADB ABE \Ð=°-°-Ð-°-Ð22ADB ABE=Ð+Ð2DBE=Ð【点睛】本题考查三角形的基本知识、全等三角形的性质和判定以及等腰三角形的性质与判定．综合性较强．18．在平行四边形ABCD 中，AB CD ^于E ，CF AD ^于F ，H 为AD 上一动点，连接CH ，CH 交AE 于G ，且4AE CD ==．（1）如图1，若60B Ð=°，求CF 、AF 的长；（2）如图2，当FH FD =时，求证：CG ED AG =+；（3）如图3，若60B Ð=°，点H 是直线AD 上任一点，将线段CH 绕C 点逆时针旋转60°，得到线段CH ¢AH ¢的最小值_____．【答案】（1）CF =，2AF =；（2）见解析；（3）4-．【分析】（1）由平行四边形性质可得60B D Ð=Ð=°，利用30°直角三角形性质开得122DF DC ==，根据勾股定理CF =，设DE a =，则2AD a =，根据勾股定理()22242a a +=，解得a =（2）方法1补短：如图3，延长GA 到M 使AM DE =，连接MB 、MC ，由平行四边形ABCD 性质，可得AB ∥CD ，AB =CD ，可证ADE BMA ≌△△（SAS ），可得BM AD =，D BMA Ð=Ð，由CF 垂直平分DH ，CH CD =，可证BM BC =，再证GM GC =即可；方法2截长：如图4，过点B 作BN CH ^于点N ，连接BG ，先证CFH CFD≌△△（SAS ），再证BNC AED ≌△△（AAS ），最后证Rt Rt ABG NBG ≌△△（HL ），可得AG GN=即可，（3）在DA 上截取DP DC =，连接CP 、PH ¢，先证CDP V 是等边三角形，可得60DCP HCH ¢Ð=Ð=°，CD CP =，可证CDH CPH ¢≌△△（SAS ），可证PH AE ¢^，设PH′交AE 与Q ，点H′在射线PQ 上运动，当点H′运动到点Q 是AH′最短由30DAE Ð=°，先求122QP AP ==，由勾股定理4AQ =-【详解】（1）由平行四边形性质可得60B D Ð=Ð=°，在Rt CFD V 中，60D Ð=°，30FCD Ð=°，4CD =，∴122DF DC ==，根据勾股定理CF ===在Rt AED △中，60D Ð=°，30DAE Ð=°，4AE =，设DE a =，则2AD a =，根据勾股定理222AE DE AD +=，即()22242a a +=，解得a ，∴AD =2AF AD DF =-=；（2）方法1补短：如图3，延长GA 到M 使AM DE =，连接MB 、MC ，∵四边形ABCD 为平行四边形，∴AB ∥CD ，AB =CD ，∵AE CD^∴AE ⊥AB ，∴90AED BAM Ð=Ð=°，在△ADE 和△BMA 中，AE AB AED BAM DE MA =ìïÐ=Ðíï=î，∴ADE BMA ≌△△（SAS ），∴BM AD =，D BMA Ð=Ð，∵CF DH ^，DF HF =，∴CF 垂直平分DH ，∴CH CD =，∴D DHC Ð=Ð，∵D BMA Ð=Ð，DHC BCH Ð=Ð，∴BMA BCH Ð=Ð，∵AD BC =，∴BM BC =，∴BMC BCM Ð=Ð，∴GMC GCM Ð=Ð，∴GM GC =，∴GC GM AG AM AG DE ==+=+．方法2截长：如图4，过点B 作BN CH ^于点N ，连接BG ，∵CF ⊥AD ，∴CFH CFD Ð=Ð，在△CFH 和△CFD 中，FH FD CFH CFDCF CF =ìïÐ=Ðíï=î∴CFH CFD ≌△△（SAS ），∴CHF D Ð=Ð，∵//AD BC ，∴CHF HCB Ð=Ð，在△BNC 和△AED 中，NCB D BNC AEDBC AD Ð=ÐìïÐ=Ðíï=î∴BNC AED ≌△△（AAS ），∴CN DE =，BN AE =，∵AE CD AB ==，∴BN AB =，∵//AB CD ，AE CD ^于E ，∴90BAG Ð=°=∠BNG ，在Rt △ABG 和Rt △NBG 中BG BG AB NB=ìí=î∴Rt Rt ABG NBG ≌△△（HL ），∴AG GN =，∴CG GN NC AG DE =+=+．（3）在DA 上截取DP DC =，连接CP 、PH ¢，∵60B D Ð=Ð=°，∴CDP V 是等边三角形，∴60DCP HCH ¢Ð=Ð=°，CD CP =，∴DCH PCH ¢Ð=Ð，在△CDH 和△CPH′中，CD CH DCH PCH CH CH =ìïÐ=Ðíï=¢¢î¢∴CDH CPH ¢≌△△（SAS ），∴60CPH D ¢Ð=Ð=°，∴CPH PCD ¢Ð=Ð，∴//PH CD ¢，∵AE CD ^，∴PH AE ¢^，设PH′交AE 与Q ，点H′在射线PQ 上运动，当点H ′运动到点Q 是AH′最短∵30DAE Ð=°由（1）得：AD ，4CD DP ==，4AP =，∴12QP AP ==，42AQ==-，∴AH ¢的最小值为4-．故答案为：4-．【点睛】本题考查平行四边形性质，30度直角三角形性质，勾股定理，三角形全等判定与性质，线段垂直平分线性质，等边三角形判定与性质，垂线段最短，掌握平行四边形性质，30度直角三角形性质，勾股定理，三角形全等判定与性质，线段垂直平分线性质，等边三角形判定与性质，垂线段最短是解题关键．19．问题提出，如图1所示，等边△ABC内接于⊙O，点P是¶AB上的任意一点，连结PA，PB，PC．线段PA、PB、PC满足怎样的数量关系?（尝试解决）为了解决这个问题，小明给出这样种解题思路：发现存在条件CA=CB，∠ACB=60°，从而将CP绕点逆时针旋转60°交PB延长线于点M，从而证明△PAC≌△MBC，请你完成余下思考，并直接写出答案：PA、PB、PC的数量关系是；（自主探索）如图2所示，把原问题中的“等边△ABC”改成“正方形ABCD”，其余条件不变，①PC与PA，PB有怎样的数量关系?请说明理由：②PC+PD与PA，PB的数量关系是．（直接写出结果）（灵活应用）把原问题中的“等边△ABC”改成“正五边形ABCDE”，其余条件不变，则PC+PD+PE与PA+PB的数量关系是．（直接写出结果）【答案】【尝试解决】PA+PB=PC；【自主探索】①PC PA=；理由见解析；②PC PD PE PA PB++=+．PC PD PA PB+=+；【灵活应用】2)()1)()【分析】尝试解决：利用旋转性质证明△PAC≌△MBC，得到PA=BM，得到PM等于PB与PA 的和，再证明△PCM是等边三角形，得到PM等于PC，即可得到结果；自主探索：①在PC上截取QC=PA，证出△CBQ全等于△ABP，得到△PBQ是等腰直角三角形，PQ等于PB②同①方法，即可得到PD与PA和PB的关系，即可求出PC+PD与PA和PB的关系；灵活应用：类比（自主探索）中的方法证明PC与PA和PB的关系，再用同样的方法证明PE与PA和PB的关系，构造△CDM全等于△CBP，得到PD与PC的关系，进一步得到PD与PA和PB的关系，最终求出PD+PE+PC的和即可得到与PA和PB的关系．【详解】尝试解决：PA+PB=PC；证明：因为∠ACP+∠PCB=60°，∠MCB+∠PCB=60°，∴∠ACP=∠MCB，又∵CP=CM，AC=MC，∴△ACP≌△BCM，所以PA=BM，∠CBM=∠CAP，∵四边形APBC内接于圆O，∴∠CAP+∠CBP=180°，∴∠CBM+∠CBP=180° ，∴P、B、M三点共线，∴△PCM 是等边三角形，∴PM=PC ，∴PC=PM=PB+BM=PB+PA ；自主探索：①PC 与PA 、PB 的数量关系为PC PA =；理由：截取CQ=PA ，，如图，∵四边形ABCD 是正方形，∴BC=AB ，∠ABC=∠BCD=∠CDA=∠DAB=90°，∵PA=CQ ，∠BCQ=BAP ，BC=AB∴△BCQ ≌△BAP ，∴∠CBQ=∠ABP ，BQ=BP ，∵∠CBQ+∠ABQ=90°，∴90ABP ABQ Ð+Ð=°，∴△PBQ 是等腰直角三角形，∴PB ，∴PC CQ PQ PA =+=；②1)()PC PD PA PB +=+证明：在PD 上截取DH=PB ，∵DH=PB ，∠ADH=∠ABP ，AD=AB∴△ADH ≌△ABP∴∠DAH=∠BAP ，AH=AP ，∵∠DAH+∠HAP=90°，∴∠BAP+∠HAP=90°，∴△HAP 是等腰直角三角形，∴PA ，∴PD=DH+PH=PB+，∴1)()PC PD PA PB +=+．灵活应用：2)()PC PD PE PA PB ++=++．。
专题02 完型填空-2022年江苏中考英语各大题型易错题及应对策略
完型填空-2022年江苏中考英语各大题型易错题及应对策略【明确考点】完型填空题的体裁一般为记叙文或说明文，不仅可考查学生的英语知识，还可考查学生的快速阅读能力，阅读理解能力，逻辑判断能力。
答题时要求学生做到：单句理解和语篇理解的统一，形式和内容的统一，语言知识和语言能力的统一。
单纯考查语法知识和词汇知识的题几乎不存在“完型填空”题中，绝大多数考题的四个选项在语法和词汇搭配上都很相似，有的甚至在意义上相似，或就其所处的句子或段落而言难辨是非，但结合具体的语言环境后，却只有一个最佳答案。
完型填空一般选用200词左右的短文，语言地道，简练生动，流畅易懂，情节曲折但表述清楚，夹叙夹议，寓意深刻；叙述文侧重描述故事情节变化或心理变化的过程。
完型填空题有一定的梯度和难度，既照顾了学生的能力差异，也检测了学生综合运用语言的能力；具有较好的区分度，其中的难题对优秀学生具有挑战性。
【命题特点】主要考查以下三方面内容：1. 词汇意义：主要考查在整体理解的基础上对词义的判定，考查特点是：各选项孤立地放在句中，句子在语法上均成立，多数情况下意义也成立，我们必须从整体出发，通过上下文进行猜测，推断。
2. 语法：孤立考语法知识的选项很少；但每个选项中的词或词组都以一定的语法形式在句中出现，因此有些选项既考语义、语法，又考句法，某些选项甚至明为语义辨析，实则却并存语法结构的区别。
3. 上下文理解：文章的文理脉络和特定的上下文对词与词，句与句之间的逻辑关系都有客观的限定性，因此“完型填空”也特别注意从上下文理解的角度进行考查。
【备考过程指导】1.做好词汇准备根据《考试说明》所附的词汇表，识记初中阶段已学过的词的词形，词义，词类，词的变化形式，以及与之相关的短语，句型及用法。
2. 掌握好学过的语法熟记与课文含义相关语法的典型句子，尤其要能准确应用动词的时态，语态，非谓语动词，情态动词，形容词，副词以及各类句型等。
【应试方法技巧指导】1.跳过空格、通读全文、把握大意。
2022年四川各地中考英语真题(成都广元南充德阳广元等)分项汇编专题02：完形填空(解析版)
专题02 完形填空（解析版）目录一、完型填空01（2022·四川成都·中考真题）阅读短文，从每小题所给的三个选项中选出一个最佳答案，使短文通顺，结构完整。
ASir Isaac Newton was a scientist who discovered the law of gravity（重力）. Peoplehave learned that Isaac Newton once ___16___ when he was thinking under an apple tree. ___17___ he was woken up by an apple falling on his head. He jumped up, shouting, “Gravity made the apple fall!”Did the apple really fall on him? No one ___18___ for sure. The story of the apple may have some ___19___, but people who knew him have written that it never happened. Newton had actually been studying gravity and thinking about how apples fell down from a tree and not ___20___ for many years.16．A．fell over B．fell asleep C．fell ill17．A．Suddenly B．Slowly C．Usually18．A．promises B．forgets C．knows19．A．truth B．purpose C．requirement20．A．inside B．off C．up【答案】16-20 BACAC【解析】本文介绍牛顿不是因为被苹果砸到开始研究万有引力，而是多年来一直在研究万有引力，并一直在思考苹果是如何从树上掉下来而不是向上的。
专题02 七年级下册语文能力提高综合练(解析版)
专题02 七年级下册语文能力提高综合练1．下列各组词语中加粗字注音无误的一项是（）A．羸弱(léi) 诧异(zhà) 慷慨以赴(kǎi)B．竹篾(miè) 迸溅(bèng) 鲜为人知(xiān)C．亘古(gèn) 选聘(pìn) 惊心动魄(pò)D．蓦然(mù) 告罄(qìng) 颠沛流离(pèi)【答案】C【解析】 A．诧chà。
B．鲜xiǎn。
D．蓦mò。
2．下列各组词语书写有误的一项是（）A．至死不懈任人宰割驿路梨花毛骨悚然B．风餐露宿扬扬得意疲惫不堪心有灵犀C．迫不急待海阔天空步履蹒姗畏缩不前D．语无伦次姗姗来迟海市蜃楼心有灵犀【答案】C【解析】C 急→及，姗→跚。
3．选出依次填入横线上最恰当的一组词语（）我们读所有的书，最终的目的都是读自己。
读有益的书，你会发现______的心平息下来了，有种______的感觉，你会发现你百思不得其解的______，千百年来被无数的人思考过，并且提供了各种各样的答案。
每一本在你心目中值得阅读和记住的书，都是因为其中______着未来你更期待的那个自己。
A．焦躁豁然开朗困惑蕴藏 B．焦虑豁然开朗疑惑蕴含C．焦躁茅塞顿开困惑蕴含 D．焦虑茅塞顿开疑惑蕴藏【答案】A【解析】本题考查选词填空。
“焦躁”意为“欲望受到压抑形成的神经衰弱症的症状”，多用作形容词。
“焦虑”意为“对亲人或自己生命安全、前途命运等的过度担心而产生的一种烦躁情绪”，多用作名词。
第一空后有“的”，故应填入形容词“焦躁”，可排除B、D两项。
“豁然开朗”意为“突然明白某件事”，与句中“百思”一词搭配恰当，“茅塞顿开”意为“闭塞的思路，由于得到了某种事物的启发，忽然想通了”，句中不含“启发”之意，故第二空应填入“豁然开朗”。
4.下列句子中没有语病的一句是（）A. 我校师生认真讨论和聆听了校长激动人心的学业考试动员报告。
专题02 数与式之填空题(28题)(解析版)
专题02 数与式之填空题参考答案与试题解析一．填空题（共28小题）1．（2019•上海）如果一个正方形的面积是3，那么它的边长是 √3 ．【答案】解：∵正方形的面积是3，∴它的边长是√3．故答案为：√3【点睛】本题考查了二次根式的应用，主要利用了正方形的性质和算术平方根的定义．2．（2018•上海）某商品原价为a 元，如果按原价的八折销售，那么售价是 0.8a 元．（用含字母a 的代数式表示）．【答案】解：根据题意知售价为0.8a 元，故答案为：0.8a ．【点睛】本题主要考查列代数式，解题的关键是掌握代数式书写规范与数量间的关系．3．（2018•上海）计算：（a +1）2﹣a 2＝ 2a +1 ．【答案】解：原式＝a 2+2a +1﹣a 2＝2a +1，故答案为：2a +1【点睛】此题考查了完全平方公式，熟练掌握完全平方公式是解本题的关键．4．（2017•上海）计算：2a •a 2＝ 2a 3 ．【答案】解：2a •a 2＝2×1a •a 2＝2a 3．故答案为：2a 3．【点睛】本题考查了单项式与单项式相乘，熟练掌握运算法则是解题的关键．5．（2019•浦东新区二模）52的相反数是 −52 ． 【答案】解：52的相反数是−52， 故答案为：−52．【点睛】此题主要考查了相反数，关键是掌握相反数定义．6．（2019•长宁区二模）今年春节黄金周上海共接待游客约5090000人，5090000这个数用科学记数法表示为 5.09×106 ．【答案】解：5090000＝5.09×106，故答案是：5.09×106．【点睛】此题考查科学记数法的表示方法．科学记数法的表示形式为a×10n的形式，其中1≤|a|＜10，n 为整数，表示时关键要正确确定a的值以及n的值．7．（2019•普陀区二模）月球离地球近地点的距离为363300千米，数据363300用科学记数法表示是 3.633×105．【答案】解：363300＝3.633×105，故答案为3.633×105．【点睛】本题考查了科学记数法表示较大的数，正确移动小数点位数是解题的关键．8．（2019•徐汇区二模）2018年1月，“墨子号”量子卫星实现了距离达7600000米的洲际量子密钥分发，数字7600000用科学记数法表示为7.6×106．【答案】解：7600000＝7.6×106，故答案为7.6×106【点睛】本题考查了科学记数法表示交大的数，正确移动小数点位数是解题的关键．9．（2019•松江区二模）计算：|−5|+(√2−1)0=6．【答案】解：原式＝5+1＝6．故答案为：6．【点睛】此题主要考查了实数运算，正确化简各数是解题关键．10．（2019•普陀区二模）如果a＝2，b＝﹣1，那么代数式√2a−b的值等于√5．【答案】解：∵a＝2，b＝﹣1，∴原式=√4+1=√5，故答案为：√5；【点睛】本题考查算术平方根，解题的关键熟练运用算术平方根的定义是，本题属于基础题型．11．（2019•虹口区二模）在数轴上，实数2−√5对应的点在原点的左侧．（填“左”、“右”）【答案】解：根据题意可知：2−√5＜0，∴2−√5对应的点在原点的左侧．故填：左【点睛】本题考查实数与数轴上点的对应关系，掌握了实数与数轴上的点的一一对应关系，很容易得出正确答案．12．（2019•杨浦区三模）某大型超市从生产基地以每千克a元的价格购进一种水果m千克，运输过程中重量损失了10%，超市在进价的基础上増加了30%作为售价，假定不计超市其他费用，那么售完这种水果，超市获得的利润是 0.17am 元（用含m 、a 的代数式表示）【答案】解：由题意可得，超市获得的利润是：a （1+30%）×[m （1﹣10%）]﹣am ＝0.17am （元），故答案为：0.17am ．【点睛】本题考查列代数式，解答本题的关键是明确题意，列出相应的代数式．13．（2019•徐汇区一模）如图，正方形DEFG 的边EF 在ABC 的边BC 上，顶点D 、G 分别在边AB 、AC 上．已知BC 长为40厘米，若正方形DEFG 的边长为25厘米，则ABC 的高AH 为 2003 厘米．【答案】解：设三角形ABC 的高AH 为x 厘米．由正方形DEFG 得，DG ∥EF ，即DG ∥BC ，∵AH ⊥BC ，∴AP ⊥DG ．由DG ∥BC 得△ADG ∽△ABC∴AP AH =DG BC ．∵PH ⊥BC ，DE ⊥BC ，∴PH ＝ED ，AP ＝AH ﹣PH ，∵BC 长为40厘米，若正方形DEFG 的边长为25厘米，∴x−25x =2540，解得x =2003．即AH 为2003厘米．故答案为：2003． 【点睛】本题考查了相似三角形的判定与性质．关键是由平行线得到相似三角形，利用相似三角形的性质列方程．14．（2019•徐汇区校级一模）如图，图中所有四边形都是正方形，其中左上角的n 个小正方形与右下角的1个小正方形边长相等，若最大正方形边长是最小正方形边长的m 倍，则用含n 的代数式表示m 的结果为m ＝ 2n +5 ．【答案】解：如图，过A 作AB ⊥FG 于B ，则△ABC ∽△CDE ，∴AB CD =BC DE =AC CE =2，设小正方形的边长为1，则大正方形的边长为m ，∴AB ＝m ﹣1，BF ＝n ，DE ＝1，∴BC ＝2DE ＝2，CD =12AB =12（m ﹣1），∴FG ＝FB +BC +CD +DG ＝n +2+12（m ﹣1）+1＝m ，∴m ＝2n +5，故答案为：2n +5．【点睛】本题考查了列代数式，相似三角形的性质和判定，正方形的性质，正确的作出辅助线构造相似三角形是解题的关键．15．（2019•杨浦区三模）计算：（﹣2）9÷27＝ ﹣4 ．【答案】解：原式＝﹣29÷27＝﹣22＝﹣4．故答案为：﹣4．【点睛】此题主要考查了同底数幂的除法，关键是掌握同底数幂的除法法则．16．（2019•奉贤区二模）计算：m 3÷（﹣m ）2＝ m ．【答案】解：m 3÷（﹣m ）2＝m 3÷m 2＝m ．故答案为m ．【点睛】本题考查了同底数幂相除，正确运用同底数幂相除法则是解题的关键．17．（2019•金山区二模）计算：a2÷a﹣2＝a4．【答案】解：a2÷a﹣2＝a2﹣（﹣2）＝a4，故答案为：a4．【点睛】本题考查的是同底数幂的除法、负整数指数幂，同底数幂的除法法则：底数不变，指数相减．18．（2019•崇明区二模）计算：（2x）2＝4x2．【答案】解：（2x）2＝4x2．故答案为：4x2．【点睛】此题主要考查了积的乘方运算，正确掌握运算法则是解题关键．19．（2019•静安区二模）如果关于x的二次三项式x2﹣4x+m在实数范围内不能分解因式，那么m的取值范围是m＞4．【答案】关于x的二次三项式x2﹣4x+m在实数范围内不能分解因式，就是对应的二次方程x2﹣4x+m＝0无实数根，∴△＝（﹣4）2﹣4m＝16﹣4m＜0，∴m＞4．故答案为：m＞4．【点睛】本题考查二次三项式的因式分解问题，可转化为对应的二次方程的实数根的情况，属于比较简单的问题．20．（2019•金山区二模）因式分解：a3+2a＝a（a2+2）．【答案】解：a3+2a＝a（a2+2），故答案为a（a2+2）．【点睛】本题考查了因式分解，正确提取公因式是解题的关键．21．（2019•浦东新区二模）分解因式：a2﹣2ab+b2﹣4＝（a﹣b+2）（a﹣b﹣2）．【答案】解：a2﹣2ab+b2﹣4＝（a﹣b）2﹣4＝（a﹣b+2）（a﹣b﹣2）．故答案为：（a﹣b+2）（a﹣b﹣2）．【点睛】此题主要考查了分组分解法因式分解，正确分组得出是解题关键．22．（2019•闵行区二模）分解因式：x2﹣9x＝x（x﹣9）．【答案】解：原式＝x•x﹣9•x＝x（x﹣9），故答案为：x （x ﹣9）．【点睛】本题考查了提公因式法因式分解的知识，解题的关键是首先确定多项式各项的公因式，然后提取出来．23．（2019•徐汇区二模）在实数范围内分解因式x 3﹣4x 的结果为 x （x +2）（x ﹣2） ．【答案】解：x 3﹣4x ＝x （x 2﹣4）＝x （x +2）（x ﹣2）．故答案为：x （x +2）（x ﹣2）．【点睛】本题主要考查了因式分解的方法，正确运用各种方法是解题的关键．24．（2019•长宁区二模）计算：(12)−2−23÷24= 312 ．【答案】解：原式＝4﹣2﹣1 ＝4−12＝312． 故答案为：312． 【点睛】此题主要考查了负指数幂的性质以及有理数的混合运算法则，正确掌握相关运算法则是解题关键．25．（2019•静安区二模）如果√x x有意义，那么x 的取值范围是 x ＞0 ． 【答案】解：由题意可知：{x ≥0x ≠0， 解得：x ＞0，故答案为：x ＞0．【点睛】本题考查二次根式，解题的关键是熟练运用二次根式有意义的条件以及分式有意义的条件，本题属于基础题型．26．（2019•金山区二模）化简：√a 3b 24（b ≥0）的结果是 ab √a 2． 【答案】解：√a 3b 24=ab √a 2，故答案为：ab √a 2．【点睛】本题主要考查的是二次根式的性质与化简，熟练掌握相关知识是解题的关键．27．（2019•杨浦区三模）计算：√18+√32= 7√2 ．【答案】解：原式＝3√2+4√2=7√2．故答案为：7√2【点睛】此题考查了二次根式的加减法，熟练掌握法则是解本题的关键．28．（2019•青浦区二模）如果二次根式√x−3有意义，那么x的取值范围是x≥3．【答案】解：∵二次根式√x−3有意义，∴x﹣3≥0，∴x≥3．故答案为：x≥3．【点睛】此题考查了二次根式有意义的条件，要明确，当函数表达式是二次根式时，被开方数非负.。