题目内容
计算
(1)(-1)2012+(-
)-2-(3.14-π)0;
(2)(3a2)2•b2÷(8a3b);
(3)[(x+y)2-y(2x+y)-8x]÷2x;
(4)
•(
)2-
÷
.
(1)(-1)2012+(-
| 1 |
| 2 |
(2)(3a2)2•b2÷(8a3b);
(3)[(x+y)2-y(2x+y)-8x]÷2x;
(4)
| x+1 |
| x |
| 2x |
| x+1 |
| 1 |
| x2-1 |
| 2 |
| x-1 |
考点:整式的混合运算,分式的混合运算,零指数幂,负整数指数幂
专题:
分析:(1)利用整式的混合运算顺序求解即可.
(2)利用整式的混合运算顺序求解即可.
(3)利用整式的混合运算顺序求解即可.
(4)利用分式的混合运算顺序求解即可.
(2)利用整式的混合运算顺序求解即可.
(3)利用整式的混合运算顺序求解即可.
(4)利用分式的混合运算顺序求解即可.
解答:解:(1)(-1)2012+(-
)-2-(3.14-π)0;
=1+4-1,
=4,
(2)(3a2)2•b2÷(8a3b)
=9a4b2÷(8a3b),
=
ab,
(3)[(x+y)2-y(2x+y)-8x]÷2x;
=[x2+2xy+y2-2xy-y2-8x]÷2x,
=(x2-8x)÷2x,
=
x-4,
(4)
•(
)2-
÷
.
=
•
-
•
,
=
-
,
=
.
| 1 |
| 2 |
=1+4-1,
=4,
(2)(3a2)2•b2÷(8a3b)
=9a4b2÷(8a3b),
=
| 9 |
| 8 |
(3)[(x+y)2-y(2x+y)-8x]÷2x;
=[x2+2xy+y2-2xy-y2-8x]÷2x,
=(x2-8x)÷2x,
=
| 1 |
| 2 |
(4)
| x+1 |
| x |
| 2x |
| x+1 |
| 1 |
| x2-1 |
| 2 |
| x-1 |
=
| x+1 |
| x |
| 4x2 |
| (x+1)2 |
| 1 |
| (x+1)(x-1) |
| x-1 |
| 2 |
=
| 4x |
| x+1 |
| 1 |
| 2(x+1) |
=
| 8x-1 |
| 2x+2 |
点评:本题主要考查了整式的混合运算,分式的混合运算,零指数幂及负整数的指数幂,解题的关键是熟记整式的混合运算,分式的混合运算,零指数幂及负整数的指数幂法则.
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