x^2-y^2-z^2-2yz-2x+1因式分解
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x^2-y^2-z^2-2yz-2x+1因式分解
x^2-y^2-z^2-2yz-2x+1因式分解
x^2-y^2-z^2-2yz-2x+1因式分解
x^2-y^2-z^2-2yz-2x+1
=x^2-2x+1-(y^2+z^2+2yz)
=(x-1)^2-(y+z)^2
=[(x-1)+(y+z)][(x-1)-(y+z)]
=(x+y+z-1)(x-y-z-1)
x^2-y^2-z^2-2yz-2x+1
=(x^2-2x+1)-(y^2+z^2+2yz)
=(x-1)^2-(y+z)^2
=(x+y+z-1)(x-y-z-1)
(x^2-2x+1)-(y^2+z^2+2yz)
=(x-1)^2-(y+z)^2
=[(x-1)+(y+z)][(x-1)-(y+z)]
=(x+y+z-1)(x-y-z-1)
x^2-y^2-z^2-2yz-2x+1
=x^2-(y^2+z^2+2yz)-2x+1
=x^2-2x+1-(y+z)^2
=(x-1)^2-(y+z)^2
=(x-1+y+z)(x-1-y-z)
(2X+Z-Y)/(X^2-XY+XZ-YZ)-(Y-Z)/(X^2-XY-XZ+YZ)
因式分解 (x+y+z)^2+yz(y+z)+xyz
化简(2x-y-z/x^2-xy-xz+yz)+(2y-x-z/y^2-xy-yz+xz)+(2x-x-y/z^2-xz-yz+xy)
化简x^2-yz/[x^2-(y+z)x+yz]+y^2-zx/[y^2-(z+x)y+zx]+z^2-xy/[z^2-(x+y)z+xy]
若|x-3|+|y+2|+|2z+1|=0,求(xy-yz)(y-x+z)
求(2X+Z-Y)/(X^2-XY+XZ-YZ)-(2X+Y+Z)/(X^2+XY+XZ+YZ)
f(x,y,z)=yz+xz使得,y^2+z^2=1,yz=3,求f最大值
(x^2+yz)/(x^2+x(y-z)-yz)+(y^2-xz)/(y^2+y(x+z)+zx)+(z^2+yx)/(x^2-z(x-y)-yx)最后一项应为:(z^2-xy)/(z^2+z(y-x)-xy)
(x+y-xy)/(x+y+2xy)=(y+z-2yz)/(y+z+3yz)=(z+x-3zx)/(z+x+4zx),且2/x=3/y=1/z,则xyz=?
已知(x+y-xy)/(x+y+2xy)=(y+z-2yz)/(y+z+3yz)=(z+x-3zx)/(z+x+4zx),且2/x=3/y-1/z,则xyz=?
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