已知x+y=1,则2x^2=4xy+2y^2的值为

已知x+y=1,则2x^2=4xy+2y^2的值为

2x^2+4xy+2y^2
=2(x^2+2xy+y^2)
=2(x+y)^2
=2*1^2
=2 所以最终答案是2

2x^2+4xy+2y^2
=2(x^2+2xy+y^2)
=2(x+y)^2
=2*1^2
=2

你那是个等式 是让求xy的值的吧 除以2得 x方=2xy+y方 加x方得 2x方=x方+2xy+y方 2x方=(x+y)方 所以2x方=1 x=+-2分之跟2 代入求得y=1+-2分之跟2

2x^2+4xy+2y^2
=2(x^2+2xy+y^2)
=2(x+y)^2
=2*1^2
=2.