已知（x+y）²-2x-2y+1=0,求2x²+4xy+2y²

已知（x+y）²-2x-2y+1=0,求2x²+4xy+2y²
已知（x+y）²-2x-2y+1=0,求2x²+4xy+2y²

已知（x+y）²-2x-2y+1=0,求2x²+4xy+2y²
∵（x+y）²-2x-2y+1=0
∴x+y=1
∴2x²+4xy+2y²=2(x+y)^2=2×1=2

(x+y)(x+y)-2(x+y)+1=0；设(x+y)=t 则t^2-2t+1=0；t=1 即(x+y)=1所以2x^2+4xy+2y^2=2(x+y)^2=2*1=2