题目内容
计算
(1)2(x-y)2-(2x+y)(-y+2x);
(2)
+|-4|+(-1)0-(
)-1
(3)(a-
)÷
×
.
(1)2(x-y)2-(2x+y)(-y+2x);
(2)
| 9 |
| 1 |
| 2 |
(3)(a-
| a |
| a+1 |
| a2-2a |
| a2-4 |
| 1 |
| a+2 |
考点:分式的混合运算,整式的混合运算,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)利用零指数幂及负整数指数幂的法则求解;
(3)利用分式的混合运算顺序求解即可.
(2)利用零指数幂及负整数指数幂的法则求解;
(3)利用分式的混合运算顺序求解即可.
解答:解:(1)2(x-y)2-(2x+y)(-y+2x)
=2(x2-2xy+y2)-(4x2-y2),
=-2x2-4xy+3y2;
(2)
+|-4|+(-1)0-(
)-1
=3+4+1-2,
=6;
(3)(a-
)÷
×
=
•
•
,
=
.
=2(x2-2xy+y2)-(4x2-y2),
=-2x2-4xy+3y2;
(2)
| 9 |
| 1 |
| 2 |
=3+4+1-2,
=6;
(3)(a-
| a |
| a+1 |
| a2-2a |
| a2-4 |
| 1 |
| a+2 |
=
| a2 |
| a+1 |
| (a+2)(a-2) |
| a(a-2) |
| 1 |
| a+2 |
=
| a |
| a+1 |
点评:本题主要考查了分式的混合运算,整式的混合运算,零指数幂及负整数指数幂,解题的关键是熟记分式的混合运算,整式的混合运算,零指数幂及负整数指数幂的法则.
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