因为（x+1)^2+|y+1|+|z|=0
又因为（x+1)^2≥0,|y+1|≥0,|z|≥0
所以x=-1,y=-1,z=0
所以 A-(2B-3C)
=2x^2-xyz-[2(y^2-z^2+xyz)-3(-x^2+2y^2-xyz)]
=2x^2-xyz-2y^2+2z^2-2xyz-3x^2+6y^2-3xyz
=-x^2+4y^2+2z^2-6xyz
=-1+4+0-0
=3