题目内容
已知(a+2)2+|b-| 1 | 2 |
分析:先根据(a+2)2+|b-
|=0,求出a,b.再根据整式的加减、去括号法则化简,代入求值即可.
| 1 |
| 2 |
解答:解:∵(a+2)2+|b-
|=0,
则a+2=0,a=-2;
b-
=0,b=
.
则5a2b-[2a2b-(ab2-2a2b)-4]-2ab2
=5a2b-[2a2b-ab2+2a2b-4]-2ab2
=5a2b-2a2b+ab2-2a2b+4-2ab2
=a2b-ab2+4
=2+
+4
=
.
| 1 |
| 2 |
则a+2=0,a=-2;
b-
| 1 |
| 2 |
| 1 |
| 2 |
则5a2b-[2a2b-(ab2-2a2b)-4]-2ab2
=5a2b-[2a2b-ab2+2a2b-4]-2ab2
=5a2b-2a2b+ab2-2a2b+4-2ab2
=a2b-ab2+4
=2+
| 1 |
| 2 |
=
| 13 |
| 2 |
点评:考查了非负数的和为0,非负数都为0.解决此类题目的关键是熟练运用多项式的加减运算、去括号法则.括号前添负号,括号里的各项要变号.先化简再代入可以简便计算.
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