题目内容

已知:(x-
12
)2+|y+3|=0
,求:3x2y-2x2y+[9x2y-(6x2y+4x2)]-(3x2y-8x2)的值.
分析:(x-
1
2
)2+|y+3|=0
,据非负数≥0,即任意数的偶次方或绝对值都是非负数,故只能x-
1
2
=0,和y+3=0;
将3x2y-2x2y+[9x2y-(6x2y+4x2)]-(3x2y-8x2)去括号,化简得x2y+4x2,问题可求.
解答:解:由题意,∵(x-
1
2
)2+|y+3|=0

∴x-
1
2
=0,y+3=0,
即x=
1
2
,y=-3;
∴3x2y-2x2y+[9x2y-(6x2y+4x2)]-(3x2y-8x2),
=3x2y-2x2y+9x2y-6x2y-4x2-3x2y+8x2
=x2y+4x2
=x2(y+4),
=(
1
2
2×(-3+4),
=
1
4
点评:本题综合考查了非负数的性质和化简求值,正确解答的关键是掌握:非负数≥0,这个知识点.
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