题目

已知3(x-1/2)^2+4丨y+1丨=0,求3x^2y^2+2xy-7x^2y^2-3/2xy+4x^2y^2的值

提问时间：2020-10-13

答案

∵3(x-1/2)^2+4丨y+1丨=0,
∴x-1/2=0 y+1=0
∴x=1/2 y=-1
∴3x^2y^2+2xy-7x^2y^2-3/2xy+4x^2y^2
=(3-7+4)x^2y^2+(2-3/2)xy
=1/2xy
=1/2×1/2×(-1)
=-1/4