已知|x+2|+(y-1/2)^2=0,求代数式2x^2y-2[xy^2-3(xy^2-2x^2
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已知|x+2|+(y-1/2)^2=0,求代数式2x^2y-2[xy^2-3(xy^2-2x^2
+1)]的值
+1)]的值
∵|x+2|+(y-1/2)²=0
∴x+2=0、y-1/2=0
x=-2、y=1/2
2x²y-2[xy²-3(xy²-2x²+1)]
=2x²y-2[xy²-3xy²+6x²-3]
=2x²y-2xy²+6xy²-12x²+6
=2x²y+4xy²-12x²+6
=2×(-2)²×1/2+4×(-2)×(1/2)²-12×(-2)²+6
=4-2-48+6
=-40
![](http://img.wesiedu.com/upload/3/d0/3d09b93f7442ef055f95d4bedb591df6.jpg)
∴x+2=0、y-1/2=0
x=-2、y=1/2
2x²y-2[xy²-3(xy²-2x²+1)]
=2x²y-2[xy²-3xy²+6x²-3]
=2x²y-2xy²+6xy²-12x²+6
=2x²y+4xy²-12x²+6
=2×(-2)²×1/2+4×(-2)×(1/2)²-12×(-2)²+6
=4-2-48+6
=-40
![](http://img.wesiedu.com/upload/3/d0/3d09b93f7442ef055f95d4bedb591df6.jpg)
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