题目内容
计算:
(1)(-2)-2+(π-3.14)0-
+(-
)-1;
(2)(x+y-
)(x-y+
).
(1)(-2)-2+(π-3.14)0-
| 3 | 27 |
| 1 |
| 3 |
(2)(x+y-
| 4xy |
| x+y |
| 4xy |
| x-y |
分析:(1)根据负整数指数幂、零指数幂和根式的化简知识点分别进行计算,再把所得的结果合并即可;
(2)根据(a+b)(a-b)=a2-b2一步步的进行分解,直到不能分解为止,即可得出答案.
(2)根据(a+b)(a-b)=a2-b2一步步的进行分解,直到不能分解为止,即可得出答案.
解答:解:(1)(-2)-2+(π-3.14)0-
+(-
)-1
=
+1-3-3
=-4
;
(2)(x+y-
)(x-y+
)
=[x+(y-
)][x-(y-
)]
=x2-(y-
)2
=(x+y-
)(x-y-
).
| 3 | 27 |
| 1 |
| 3 |
=
| 1 |
| 4 |
=-4
| 3 |
| 4 |
(2)(x+y-
| 4xy |
| x+y |
| 4xy |
| x-y |
=[x+(y-
| 4xy |
| x+y |
| 4xy |
| x+y |
=x2-(y-
| 4xy |
| x+y |
=(x+y-
| 4xy |
| x+y |
| 4xy |
| x+y |
点评:此题考查了实数的运算以及分式的化简,掌握负整数指数幂、零指数幂和根式的化简以及分式化简的步骤是解题的关键.
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