试题

题目:
计算:
(1)(x2-y2)·
xy
x2+2xy+y2
÷
x2-xy
x+y

(2)
2x
x2-9
+
1
3-x
-
2
x2+6x+9

答案
解:(1)原式=(x+y)(x-y)×
xy
(x+y)2
×
x+y
x(x-y)

=y;
(2)原式=
2x
(x+3)(x-3)
-
1
x-3
-
2
(x+3)2

=
2x(x+3)
(x-3)(x+3)2
-
(x+3)2
(x-3)(x+3)2
-
2(x-3)
(x-3)(x+3)2

=
2x2+6x-x2-6x-9-2x+6
(x-3)(x+3) 2

=
x+1
(x+3) 2

解:(1)原式=(x+y)(x-y)×
xy
(x+y)2
×
x+y
x(x-y)

=y;
(2)原式=
2x
(x+3)(x-3)
-
1
x-3
-
2
(x+3)2

=
2x(x+3)
(x-3)(x+3)2
-
(x+3)2
(x-3)(x+3)2
-
2(x-3)
(x-3)(x+3)2

=
2x2+6x-x2-6x-9-2x+6
(x-3)(x+3) 2

=
x+1
(x+3) 2
考点梳理
分式的混合运算.
(1)先进行因式分解,再约分即可;
(2)先进行因式分解,再通分,按同分母的分式进行计算即可.
本题考查了分式的混合运算,是基础知识要熟练掌握.
计算题.
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