已知:(x+2)2+|y-12|=0,求2(xy2+x2y)-[2xy2-3(1-x2y)]-2的值.-数学

题目简介

已知:(x+2)2+|y-12|=0,求2(xy2+x2y)-[2xy2-3(1-x2y)]-2的值.-数学

题目详情

已知:(x+2)2+|y-
1
2
|=0
,求2(xy2+x2y)-[2xy2-3(1-x2y)]-2的值.
题型:解答题难度:中档来源:不详

答案

原式=2xy2+2x2y-[2xy2-3+x2y]-2=2xy2+2x2y-2xy2+3-3x2y-2
=(2-2)xy2+(2-3)x2y+(3-2)=-x2y+1;
∵(x+2)2≥0,|y-class="stub"1
2
|≥0

又∵(x+2)2+|y-class="stub"1
2
|=0

∴x=-2,y=class="stub"1
2

∴原式=-(-2)2×class="stub"1
2
+1

=-1.

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