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.

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.
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.

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

x^2+y^2+5=2x+4y
于是就有
x²-2x+1+y²-4y+4=0
分别配方就可以得到
（x-1)²+（y-2)²=0
于是就有
x-1=0和y-2=0
解得x=1,y=2
于是
[2x^2-(x+y)(x-y)][(x+y-1)(x-y+1)+1-2y]
=[2-(1+2)(1-2)][(1+2-1)(1-2+1)+1-4]
=5×（-3）
=-15

X=1，Y=2（配方，然后利用非负性来做）
代入计算

[2x^2-(x+y)(x-y)][(x+y-1)(x-y+1)+1-2y]
=[2-(1+2)(1-2)][(1+2-1)(1-2+1)+1-4]
=5×（-3）
=-15