Question

已知x²+y²-4x+y+4又1/4=0，求y的-x次方+3xy的值

Answer 1

x²+y²-4x+y+4+1/4=0
x²-4x+4+y²+y+1/4=0
(x-2)²+(y+1/2)²=0
则：x-2=0，y+1/2=0
得：x=2，y=-1/2
所以，y的-x次方+3xy=4-3=1

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Answer 2

x²+y²-4x+y+4又1/4=0
(x²-4x+4)+(y²+y+1/4)=0
(x-2)²+(y+1/2)²=0
平方项恒非负，两平方项之和=0，两平方项分别=0
x-2=0 x=2
y+1/2=0 y=-1/2
y^(-x)+3xy
=(-1/2)^(-2)+3×2×(-1/2)
=4-3
=1

Answer 3

x²+y²-4x+y+4又1/4=0
(x-2)²+(y+1/2)²=0

x-2=0
y+1/2=0
x=2
y=-1/2

y的-x次方+3xy
=(-1/2)^-2+3*2*(-1/2)
=4-3
=1

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