问题
解答题
计算: (1)(
(2)(-
(3)(
(4)
|
答案
(1)原式=
•x6y3 -z6
=-z x3
;x3y3 z5
(2)原式=
•(-a4 b2
•b6 a3
)=-a5;a4 b4
(3)原式=
•(a-b)2 a2b2
•(a+b)(a-b)=a3 (a-b)3
;a(a+b) b2
(4)原式=-
•(m+4)(m-4) (m+4)2
•2(m+4) m-4
=-m-2 m+2
.2m-4 m+2