已知x²+y²+5=2x+4y,求代数式（2x²-（x+y)(x-y))x((x+y-1)(x-y+1)+(1-2y))

已知x²+y²+5=2x+4y所以(x-1)²+(y-2)²=0故x=1,y=2所以（2x²-（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))=（2+3)×(-3)=-15