（1）(x+y+z)(x-y-z)= (2)2012^2-2014×201=

已知 x/(y+z)+y/(z+x)+z/(x+y)=1求 (x*x)/(y+z)+(y*y)/(x+z)+(z*z)/(x+y)=? 已知实数X.Y.Z满足(Y+Z)分之X+(Z+X)分之Y+(X+Y)分之Z=1,则(Y+Z)分之X平方+(Z+X)分之Y平方+(X+Y)分之Z平方的值为（ ） (x+y-z)(x-y+z)= x分之y+z=y分之z+x=z分之x+y(x+y+z不等于0）,求x+y+z分之x+y-z 分式加减法：已知x+y/z=x+z/y=y+z/x(x+y+z≠0）,求x+y-z/x+y+z 已知x+y/z=x+z/y=y+z/x(x+y+z≠0）,求x+y-z/x+y+z的步骤 已知x+y/z=x+z/y=y+z/x(x+y+z≠0）,求x+y-z/x+y+z的步骤 1.设有比例式：x/(y+z)=y/(x+z)=z/(x+y),有比例性质,得x/(y+z)=y/(x+z)=z/(x+y)=（x+y+z)/[(y+z)+(x+z)+(x+y)]=0.5x/(y+z)=y/(x+z)=(x-y）/[（y+z）-（x+z)]=-1,由此可得0.5=-1试分析推理产生错误的原因 (x-y-z)*( )=x²-（y-z）² 已知x,y,z>0,xyz(x+y+z)=1,求证（x+y）(x+z)>=2 （x+y+z）*（-x+y+z）*（x-y+z）*（x+y-z）= X+Y+Z=? （x+y-z）（x-y+z）-（x+y+z）（x-y-z） x/y=(x+z)/(y+z)y/z=(x+y)/(x+z) （x+y-z）*（x+y-z)-(x-y+z)*(x-y-z)等于多少 （x+y+z）(x-y+z)+(y-x+z)(y-x-z)求过程 （x-y+z)(-x+y+z)=[z+( )][ ]=z²-（ ）² {x+y-z=11 y+z-x=5 z+x-y=1