问题 解答题
计算:
(1)
1
2
-1
+
3
3
-
6
)+
8

(2)(a2
n
m
-
ab
m
mn
+
n
m
m
n
)÷a2b2
n
m
(m>0,n>0).
答案

(1)原式=

2
+1
(
2
-1)(
2
+1)
+(
9
-3
2
)+2
2

=

2
+1+3-3
2
+2
2

=4;

(2)原式=(a2

mn
m
-
ab
m
mn
+
n
m
mn
n
)÷(a2b2
mn
m

=

mn
a2-ab
m
+
1
m
)×
1
a 2b 2
×
m
mn

=

a2-ab+1
a2b2

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