若x+y=2,且(x+2)(y+2)=5,则计算xy=_;3x^2+xy+3y^2=_有过程
问题描述：
若x+y=2,且(x+2)(y+2)=5,则计算xy=_;3x^2+xy+3y^2=_
有过程
答
x+y=2,
(x+2)(y+2)=5
展开得：
xy+2(x+y)+4=5
xy+2×2+4=5
xy=-3
3x^2+xy+3y^2
=3(x²+2xy+y²)-5xy
=3(x+y)²-5xy
=3×2²-5×(-3)
=12+15
=27
答
(x+2)(y+2)
=xy+2x+2y+4
=xy+4+4
=5
xy=5-8=-3
3x^2+xy+3y^2
=3(x+y)^2-5xy
=3*4-5*(-3)
=27