Question

因式分解..公式法

Answer 1

1）
原式=[2(a+b)]²-[5(a-c)]²
=(2a+2b+5a-5c)(2a+2b-5a+5c)
=(7a+2b-5c)(-3a+2b+5c)

2)原式=1/2 （a²-4）
=1/2(a+2)(a-2)

Answer 2

（1）4（a+b）²-25（a-c）²
=[2(a+b)]²-[5(a-c)]²
=[2(a+b)+5(a-c)][2(a+b)-5(a-c)]
=(7a+2b-5c)(-3a+2b+5c)
（2）1/2a²-2
=1/2(a+2)（a-2）
希望能帮到你，望采纳，谢谢！

Answer 3

(1)原式=[2(a+b)]²-[5(a-c)]²=[2(a+b)+5(a-c)][2(a+b)-5(a-c)]=(7a+2b-5c)(-3a+2b+5c)
(2)原式=1/2(a²-4)=1/2(a²-2²)=(a+2)(a-2)/2

Answer 4

Answer 5

(1) 解：
原式 = [ 2(a+b)]^2 - [ 5(a-c)]^2
= [ 2(a+b)]+[ 5(a-c)] [ 2(a+b)] - [ 5(a-c)]
=( 7a+2b -5c ) (-3a+2b+ 5c)
(2) 解：
原式 = (1/2) ( a^2 - 2^2 ) = (1/2) ( a +2 ) ( a - 2 )