题目内容

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
(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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