题目内容

计算:(1) 
2m
3n
•(
3n
p
)2÷
mn
p2
(2)
1
x
+
1
2x
+
1
3x
(3)
a2-1
a2+2a+1
÷
a2-a
a+1

(4)
x2+1
x2-1
-
x-2
x-1
÷
x-2
x
(5) 
2a
a2-4
+
1
2-a
(6) 
x2
x-1
-x-1
分析:(1)此小题只需先乘方、再乘除、最后约分即可;
(2)此小题只需通分后将分子相加即可;
(3)此小题只需先把除法换为乘法后再约分即可;
(4)此小题应先计算第二项后,再与第一项通分计算;
(5)、(6)约可先通分,相加后再约分即可.
解答:解:(1)
2m
3n
(
3n
p
)
2
÷
mn
p2
=
2m
3n
9n2
p2
p2
mn
=6;
(2)
1
x
+
1
2x
+
1
3x
=
6+3+2
6x
=
11
6x

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

(4)
x2+1
x2-1
-
x-2
x-1
÷
x-2
x
=
x2+1
x2-1
-
x
x-1
=
x2+1-x(x+1)
x2-1
=-
1
x+1

(5)
2a
a2-4
+
1
2-a
=
2a
a2-4
-
a+2
a2-4
=
a-2
a2-4
=
1
a+2

(6)
x2
x-1
-x-1
=
x2
x-1
-
x2-1
x-1
=
1
x-1
点评:本题主要考查分式的混合运算,通分、因式分解和约分是解答的关键.
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