设A=2x^2+xy+3y^2,B=x^2-xy+2y^2,若x^2与|2x-y+2|互为相反数,求2A-3(2B-A)

A=2x^2+xy+3y^2 B=x^2-xy+2y^2
(1)
x的平方与|2x-y+2|互为相反数
x^2+|2x-y+2|=0
x=0 2x-y+2=0
y=2
2A-3(2B-A)
=2(2x^2+xy+3y^2)-3(2(x^2-xy+2y^2)-(2x^2+xy+3y^2))
=4x^2+2xy+6y^2-3(2x^2-2xy+4y^2-2x^2-xy-3y^2)
=4x^2+2xy+6y^2-3(-3xy+y^2)
=4x^2+2xy+6y^2+9xy-3y^2
=4x^2+11xy+3y^2
=4*0^2+11*0*2+3*2^2
=3*4
=12

x^2与|2x-y+2|互为相反数
x^2与|2x-y+2|都是非负数
所以x=0 y=2
所以答案是12