题目内容
计算:
(1)
;
(2)(-
)2•(-
)3÷(
)4;
(3)
-a-2;
(4)
÷(x+2-
).
(1)
| 18x4y-3 |
| 12x-1y3 |
(2)(-
| a2 |
| b |
| b2 |
| 2a |
| 1 |
| a3b2 |
(3)
| 4 |
| a+2 |
(4)
| 3-x |
| x-2 |
| 5 |
| x-2 |
分析:(1)首先利用负指数次幂的意义转化为乘法运算,然后利用单项式的乘法法则计算;
(2)首先计算乘方,然后把除法转化为乘法进行约分;
(3)首先进行通分,然后进行同分母的分式的减法即可求解;
(4)首先通分计算括号内的式子,然后把除法转化为乘法,进行约分即可.
(2)首先计算乘方,然后把除法转化为乘法进行约分;
(3)首先进行通分,然后进行同分母的分式的减法即可求解;
(4)首先通分计算括号内的式子,然后把除法转化为乘法,进行约分即可.
解答:解:(1)原式=18x4y-3×
xy-3=
;
(2)原式=
•(-
)÷(
)
=-
a13b12;
(3)原式=
-
=
=
=-
;
(4)原式=
÷
=
÷
=
•
=-
.
| 1 |
| 12 |
| 3x5 |
| 2y6 |
(2)原式=
| a4 |
| b2 |
| b6 |
| 8a3 |
| 1 |
| a12b8 |
=-
| 1 |
| 8 |
(3)原式=
| 4 |
| a+2 |
| (a+2)2 |
| a+2 |
| 4-(a+2)2 |
| a+2 |
| -a2-4a |
| a+2 |
| a2+4a |
| a+2 |
(4)原式=
| 3-x |
| x-2 |
| (x+2)(x-2)-5 |
| x-2 |
=
| 3-x |
| x-2 |
| (x+3)(x-3) |
| x-2 |
=
| 3-x |
| x-2 |
| x-2 |
| (x+3)(x-3) |
=-
| 1 |
| x+3 |
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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