问题 解答题
计算;
(1)(-
x2
y
)2•(-
y2
x
)3•(-
1
x
)4

(2)4x2y÷(
2x
-y
)2

(3)(
b2
ac
)3÷(-b6c)

(4)-
x-y
x+2y
÷
x2-y2
x2+4xy+4y2
x-y
x+2y

(5)
m2-n2
(m-n)2
(n-m)2
m2n2
÷
m+n
m
答案

(1)原式=

x4
y2
•(-
y6
x3
)•
1
x4

=-

y4
x3

(2)原式=4x2y•

y2
4x2

=y3

(3)原式=

b6
a3c3
1
-b6c

=-

1
a3c4

(4)原式=-

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

=-

x-y
x+y

(5)原式=

(m+n)(m-n)
(m-n)2
(m-n)2
m2n2
m
m+n

=

m-n
mn2

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