已知:(x+2)²+丨y-6丨=0,求2(xy²+x²y)-[2xy²-3(1-x²y)]-2的值.
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已知:(x+2)²+丨y-6丨=0,求2(xy²+x²y)-[2xy²-3(1-x²y)]-2的值.
已知:(x+2)²+丨y-6丨=0,求2(xy²+x²y)-[2xy²-3(1-x²y)]-2的值.
已知:(x+2)²+丨y-6丨=0,求2(xy²+x²y)-[2xy²-3(1-x²y)]-2的值.
(x+2)²+丨y-6丨=0
∴﹛x+2=0
y-6=0
∴x=-2,y=6
2(xy²+x²y)-[2xy²-3(1-x²y)]-2
=2xy²+2x²y-(2xy²-3+3x²y)-2
=2xy²+2x²y-2xy²+3-3x²y-2
=-x²y+1
=-(-2)²×6+1
=-23
由1式可以知道x=-2,y=6. 之后你应该能搞定了。