题目内容

计算与化简:
(1)
3-a2
a2+2a+1
+(1-
1
a+1
)÷(1+
1
a-1
)
        
(2)(
c
ab
)2•(-
b
ac
)3÷(-
b
a
)4
(1)原式=
3-a2
(a+1)2
+(
a+1
a+1
-
1
a+1
)÷(
a-1
a-1
+
1
a-1

=
3-a2
(a+1)2
+
a
a+1
a-1
a

=
3-a2
(a+1)2
+
a-1
a+1

=
3-a2
(a+1)2
+
a2-1
(a+1)2

=
2
(a+1)2

=
2
a2+2a+1

(2)原式=-
c2
a2b2
b3
a3c3
a4
b4

=-
a4b3c2
a5b6c3

=-
1
ab3c
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