题目内容
计算
(1)(-
)-1-23×0.125+(
-2012)0
(2)(
)2÷(
)2•
(3)
+
-
(4)(
-
)•
÷(
+
).
(1)(-
| 1 |
| 2 |
| 3 | -27 |
(2)(
| -a |
| b |
| 2a2 |
| 5b |
| a |
| 5b |
(3)
| m+2n |
| n-m |
| n |
| m-n |
| 2n |
| n-m |
(4)(
| x |
| x-y |
| 2y |
| x-y |
| xy |
| x-2y |
| 1 |
| x |
| 1 |
| y |
分析:(1)首先计算乘方,然后计算乘法,最后进行加减运算即可求解;
(2)首先计算乘方,然后把除法转化成乘法,然后进行乘法运算即可;
(3)首先通分,然后进行同分母的分式的加减运算即可;
(4)首先计算括号内的分式,然后把除法转化成乘法,最后进行乘法运算即可.
(2)首先计算乘方,然后把除法转化成乘法,然后进行乘法运算即可;
(3)首先通分,然后进行同分母的分式的加减运算即可;
(4)首先计算括号内的分式,然后把除法转化成乘法,最后进行乘法运算即可.
解答:解:(1)原式=-2+8×0.125+1
=-2+1+1
=0;
(2)原式=
÷
•
=
•
•
=
;
(3)原式=
-
-
=
=
=-1;
(4)原式=
•
÷
=
•
•
=
.
=-2+1+1
=0;
(2)原式=
| a2 |
| b2 |
| 4a4 |
| 25b2 |
| a |
| 5b |
=
| a2 |
| b2 |
| 25b2 |
| 4a2 |
| a |
| 5b |
=
| 5a |
| 4b |
(3)原式=
| m+2n |
| n-m |
| n |
| n-m |
| 2n |
| n-m |
=
| m+2n-n-2n |
| n-m |
=
| m-n |
| n-m |
=-1;
(4)原式=
| x-2y |
| x-y |
| xy |
| x-2y |
| x+y |
| xy |
=
| x-2y |
| x-y |
| xy |
| x-2y |
| xy |
| x+y |
=
| x2y2 |
| x2-y2 |
点评:本题考查分式的混合运算,关键是通分,合并同类项,注意混合运算的运算顺序.
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