已知(x-3)^2+丨3y+1丨=0.求3x^2y-[2xy-2(xy-3/2)+xy]+3xy^2的值
问题描述：
已知(x-3)^2+丨3y+1丨=0.求3x^2y-[2xy-2(xy-3/2)+xy]+3xy^2的值
答
已知(x-3)^2+丨3y+1丨=0.
x-3=0 x=3
3y+1=0 y=-1/3
3x^2y-[2xy-2(xy-3/2)+xy]+3xy^2
=3x^2y-2xy+2xy-3-xy+3xy^2
=3x^2y-3-xy+3xy^2
=xy(3x-1+3y)
=3×1/3×（3×3-1+3×1/3）
=1×（9-1+1）
=9
答
(x-3)^2+丨3y+1丨=0x-3=0, 3y+1=0x=3, y=-1/33x^2y-[2xy-2(xy-3/2)+xy]+3xy^2=3x^2y-2xy+2xy-3-xy+3xy^2=3x^2y-xy+3xy^2-3=3*9*(-1/3)-3*(-1/3)+3*3*(-1/3)^2-3=-9+1+1-3=-10