题目内容
计算:
(1)(2-π)0-(
)-2+(
)2011×(-4)2012
(2)(x+1)(x-1)-(x+2)2.
(1)(2-π)0-(
| 1 |
| 2 |
| 1 |
| 4 |
(2)(x+1)(x-1)-(x+2)2.
(1)原式=1-4+(-
×4)2011×(-4)=1-4+4=1;
(2)原式=x2-1-(x2+4x+4)=x2-1-x2-4x-4=-4x-5.
| 1 |
| 4 |
(2)原式=x2-1-(x2+4x+4)=x2-1-x2-4x-4=-4x-5.
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