xyz=1 x+y+z=2,x^2+y^2+z^2=3,1/(xy+z-1)+1/(xz+y-1)+1/(yz+x-1)=

x+y+z+2=xyz,x,y,z.为正实数,证明：xyz(x-1)(y-1)(z-1) 正数XYZ满足(X+Y)(X+Z)=2则XYZ(X+Y+Z)最大值 先化简再求值3xyz+2(x^2y+y^2z-xyz)-xyz+2z^2x x=1 y= -1 z=2 先化简,再求值：3xyz+2(x²y+y²z-xyz)-xyz+2z²x,其中x=1、y=-1、z=2; (x*x+2)(y*y+4)(z*z+8)=64xyz,求x,y,z 已知x,y,z>0,xyz(x+y+z)=1,求证（x+y）(x+z)>=2 3x^2y-[2x^2y-(2xyz-x^2z)-4x^2z]-xyz ,其中x=-2,y=-3,z=1 X^2Y-[-X^2Y+(XYZ-X^2Z)+XYZ]-X^2Z,其中x=-1,y=-2,z=1/3 3x²y-[2x²y-(2xyz-x²z)-4x²z]-xyz,其中x=-2 y=-3 z=1 x^2y--[-x^2y-(2XYZ-X^2Z-3x^2y)-4x^2z]-XYZ其中x=-3,y=2,z=-1x^2y--[-x^2y-(2XYZ-X^2Z-3x^2y)-4x^2z]-XYZ其中x=-3,y=2,z=-1 若xyz=1,求证 x^2/(y+z)+y^2/(z+x)+z^2/(x+y)≥3/2 xyz=1,求证：x/(1+y)+y/(1+z)+z/(1+x) >=3/2 已知x^2+y^2+z^2=1,求证x+y+z-2xyz x=2 2y=z xyz=1+9 y=? z=? (2x^3-xyz)-2(x^3-y^3+xyz)+(xyz-2y^3),其中x=-1,y=-2,z=-3. 化简求值：（2x³-xyz)-2(x³-y³+xyz)+(xyz-2y³）,x=1,y=2,z=-3 (2x³-xyz）-2（x³-y³+xyz）+（xyz-3y³） x=1 y=-2 z=-3 先化简再求值:xyz-[-x^2y+(xyz-x^2z)]-x^2z,其中x=-1,y=-2,z=1