问题
解答题
计算: (1)
(2)
(3)(
(4)(1+
|
答案
(1)原式=
-x2 x-1 x+1 1
=x2-(x+1)(x-1) x-1
=
.1 x-1
(2)原式=
-x(x+2) (x+2)(x-2) 2 x-2
=
-x x-2 2 x-2
=x-2 x-2
=1.
(3)原式=
•x(x-1)-3x(x+1) (x+1)(x-1) (x+1)(x-1) x
=-2x2-4x x
=-2x-4.
(4)原式=(
)2•(a-b+2b a-b
)2a+3b a+b
=
•(a+b)2 (a-b)2 (a+3b)2 (a+b)2
=(a+3b)2 (a-b)2
=
.a2+6ab+9b2 a2-2ab+b2