问题 解答题
计算:
(1)
x2
x-1
-x-1

(2)
x2+2x
x2-4
-
2
x-2

(3)(
x
x+1
-
3x
x-1
)÷
x
x2-1

(4)(1+
2b
a-b
)
2
(1
2b
a+b
)
2
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=(

a-b+2b
a-b
2•(
a+3b
a+b
2

=

(a+b)2
(a-b)2
(a+3b)2
(a+b)2

=

(a+3b)2
(a-b)2

=

a2+6ab+9b2
a2-2ab+b2

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