已知x平方+x平方=2x+4y求代数式[2x平方-（x+y）(x-y)][(x+y-1)(x-y+1)+1-2y]的值
问题描述：
已知x平方+x平方=2x+4y求代数式[2x平方-（x+y）(x-y)][(x+y-1)(x-y+1)+1-2y]的值
答
[2x^2-（x+y）(x-y)][(x+y-1)(x-y+1)+1-2y]=[x^2-（x^2-y^2)][x^2-(y-1)^2+1-2y]
=(x^2+y^2)(x^2-y^2+2y-1+1-2y)==(x^2+y^2)(x^2-y^2)=x^4-y^4
答
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