八年级数学上册解题技巧专题乘法公式的灵活运用(新版)华东师大版
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八年级数学上册解题技巧专题乘法公式的灵活运用(新版)华东
师大版
——计算技巧多,先观察,再计算,事半功倍
◆类型一 利用乘法公式进行简便运算
1.计算102×98的结果是( )
A .9995
B .9896
C .9996
D .9997
2.计算20162-2015×2017的结果是( )
A .-2
B .-1
C .0
D .1
3.计算:
(1)(邵阳校级月考)512=_______;
(2)82015×(-0.125)2016×(-1)2017=________.
4.运用公式简便计算:
(1)4013×3923; (2)10002
2522-2482.
5.(泰兴市校级月考)阅读下列材料:
某同学在计算3(4+1)(42+1)时,把3写成4-1后,发现可以连续运用平方差公式计
算:3(4+1)(42+1)=(4-1)(4+1)(42+1)=(42-1)(42+1)=162-1.请借鉴该同学的经验,计算下列各式的值:
(1)(2+1)(22+1)(24+1)(28+1)·…·(22004+1); (2)? ????1+12? ????1+122? ????1+124? ????1+128+1215.
◆类型二 利用乘法公式的变式求值
6.若a -b =12,且a 2-b 2=14
,则a +b 的值为( ) A .-12 B.12
C .1
D .2 7.若a -b =1,ab =2,则(a +b )2的值为( )
A .-9
B .9
C .±9 D.3
8.已知x +1x =5,那么x 2+1x 2的值为( ) A .10 B .23 C .25 D .27
9.若m +n =1,则代数式m 2-n 2+2n 的值为 .
10.阅读:已知a +b =-4,ab =3,求a 2+b 2的值.
解:∵a +b =-4,ab =3,
∴a 2+b 2=(a +b )2-2ab =(-4)2-2×3=10.
请你根据上述解题思路解答下面问题:
(1)已知a -b =-3,ab =-2,求(a +b )(a 2-b 2)的值;
(2)已知a -c -b =-10,(a -b )c =-12,求(a -b )2+c 2的值.
参考答案与解析
1.C 2.D
3.(1)2601 (2)-18
4.解:(1)原式=? ????40+13? ????40-13=402-? ??
??132=159989; (2)原式=10002(252+248)×(252-248)=10002500×4
=500. 5.解:(1)(2+1)(22+1)(24+1)(28+1)·…·(22004+1)=(2-1)(2+1)(22+1)(24+1)(28+1)·…·(22004+1)=(22-1)(22+1)(24+1)(28+1)·…·(22004+1)=(24-1)(24+1)(28+1)·…·(22004+1)=24008-1;
(2)? ????1+12? ????1+122? ????1+124? ????1+128+1215=2×? ????1-12? ????1+12? ????1+122? ????1+124? ????1+128+1215=2×? ????1-1216+1215=2-1215+1215=2. 6.B 7.B 8.B
9.1 解析:∵m +n =1,∴m 2-n 2+2n =(m +n )(m -n )+2n =m -n +2n =m +n =1.
10.解:(1)∵a -b =-3,ab =-2,∴(a +b )(a 2-b 2)=(a +b )2(a -b )=[(a -b )2+
4ab ](a -b )=[(-3)2+4×(-2)]×(-3)=-3;
(2)(a -b )2+c 2=[(a -b )-c ]2+2(a -b )c =(-10)2+2×(-12)=76.