问题 解答题
计算:
(1)
x2
x-1
-x-1

(2)
4
+(-2008)0-(
1
3
)-1+|-2|

(3)(2ab2c-3-2÷(a-2b)3
(4)
x2-y2
x
÷(x-
2xy-y2
x
)

(5)
6
x2-9
+
1
x+3
答案

(1)原式=

x2
x-1
-
x+1
1

=

x2
x-1
-
(x+1)(x-1)
x-1

=

x2-(x2-1)
x-1

=

1
x-1

(2)原式=2+1+3+2=8;

(3)原式(2ab2c-3-2•(a-2b)-3

=

1
4
a-2 b-4c6•a6b-3

=

1
4
a4b-7c6

=

a4c6
4b7

(4)原式=

(x+y)(x-y)
x
÷
x2-2xy+y2
x

=

(x+y)(x-y)
x
x
(x-y)2

=

x+y
x-y

(5)原式=

6
(x+3)(x-3)
+
x-3
(x+3)(x-3)

=

x+3
(x+3)(x-3)

=

1
x-3

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