已知(1-x)²+/y+2/=0,求4xy-【2x²-3(x²-xy+y²)】的值
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已知(1-x)²+/y+2/=0,求4xy-【2x²-3(x²-xy+y²)】的值
已知(1-x)²+/y+2/=0,求4xy-【2x²-3(x²-xy+y²)】的值
已知(1-x)²+/y+2/=0,求4xy-【2x²-3(x²-xy+y²)】的值
(1-x)²+/y+2/=0,
∴1-x=0
y+2=0
∴x=1
y=-2
4xy-【2x²-3(x²-xy+y²)】
=4xy-2x²+3x²-3xy+3y²
=x²+xy+3y²
当x=1,y=-2时
原式=1-2+12=11
即1-x=0,y+2=0
x=1,y=-2
所以原式=4xy-2x²+3x²-3xy+3y²
=x²+xy+3y²
=1-2+12
=11