题目内容
计算:
(1)|-2|-
+(
)-1+(-1)2011-(π-3)0
(2)
+
(3)(2ab2c-3)-2÷(a-2b)3
(4)
÷
.
(1)|-2|-
| 9 |
| 1 |
| 3 |
(2)
| x |
| x-y |
| y |
| y-x |
(3)(2ab2c-3)-2÷(a-2b)3
(4)
| x-2 |
| x2-1 |
| x+1 |
| x2+2x+1 |
分析:(1)根据0指数幂,负整数指数幂,正整数指数幂,绝对值,二次根式的性质进行计算;
(2)公分母为(x-y),通分化简即可;
(3)先乘方,再进行同底数幂的运算;
(4)将除法转化为乘法,分子、分母因式分解,约分.
(2)公分母为(x-y),通分化简即可;
(3)先乘方,再进行同底数幂的运算;
(4)将除法转化为乘法,分子、分母因式分解,约分.
解答:解:(1)原式=2-3+3-1-1
=0;
(2)原式=
=1;
(3)原式=2-2a-2b-4c6÷(a-6b3)
=
a-2+6b-4-3c6
=
;
(4)原式=
•
=
.
=0;
(2)原式=
| x-y |
| x-y |
=1;
(3)原式=2-2a-2b-4c6÷(a-6b3)
=
| 1 |
| 4 |
=
| a4c6 |
| 4b7 |
(4)原式=
| x-2 |
| (x+1)(x-1) |
| (x+1)2 |
| x+1 |
=
| x-2 |
| x-1 |
点评:本题主要考查分式的混合运算,实数的运算,同底数幂的运算.进行分式混合运算时,通分、因式分解和约分是解答的关键.
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