已知x^2+y^2+5=2x+4y,求代数式[2x^2-(x+y)(x-y)][(x+y-1)(x-y+1)+1-2y]

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已知x^2+y^2+5=2x+4y,求代数式[2x^2-(x+y)(x-y)][(x+y-1)(x-y+1)+1-2y]的值.

x²+y²+5=2x+4y
x²-2x+1+y²-4y+4=0
(x-1)² + (y-2)²=0
∵(x-1)²≥0且(y-2)²≥0
∴x=1,y=2
[2x²-(x+y)(x-y)][(x+y-1)(x-y+1)+1-2y]=[2x²-(x²-y²)][x²-(y-1)²+1-2y]=(x²+y²)(x²-y²)=x^4 - y^4
=1^4 - 2^4
=-15

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