问题
解答题
计算: (1)(-
(2)(
(3)
(4)
|
答案
(1)原式=-
×(-b2 2a
)×a3b3=a6 b3
a8b2;1 2
(2)原式=
•(a+b)2(a-b)2 b2
•1 a2(a+b)2
=1;a2b2 (a-b)2
(3)原式=
•a+2 (a-1)2
•(a-2)2 a+1
=(a+1)(a-1) (a+2)(a-2)
;a-2 a-1
(4)原式=
•(x+1)(x-1) (x+1)2
•4(x+1)2 2(x+1)(x-1)
=1 (x-1)2
.2 (x-1)2