题目

已知x+y=0,x+3y=1,求3x^2+12xy+13y^2 的值

提问时间：2021-03-01

答案

3x^2+12xy+13y^2
=x^2+2xy+y^2 + 2x^2+10xy+12y^2
=(x+y)^2 + 2(x+2y)(x+3y)
=0+2(x+2y)*1
=0+2(x+y + y)
=2y
=(x+3y)-(x+y)
=1-0
=1