y^2+yz+z^2=a^2,z^2+zx+x^2=b^2,x^2+xy+y^2=c^2,yz+zx+xy=0.证明:(a+b+c)(a+b-c)(a-b+c)(a-b-c)=0

因式分解 (x+y+z)^2+yz(y+z)+xyz 3（y-z）^2-(2y+z)(-z+2y)这道题我是这样做的：3（y-z）^2-(2y+z)(-z+2y)=3（y-z)^2+(-z-2y)(-z+2y)=3(y-z)^2+z^2-4y^2=3(y^2-2yz+z^2)+z^2-4y^2=3y^2-6yz+3z^2+z^2-4y^2=4z^2-6yz-y^2(a+b-c)^2=[(a+b)-c]^2=(a+b)^2-2(a+b)c+c^2=a^2+b^2-2a+2bc+c^2 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] 已知XY=2(X+Y),YZ=4(Y+Z),ZX=5(Z+X),求X,Y,Z已知XY=2(X+Y),YZ=4(Y+Z),ZX=5(Z+X),求X,Y,Z x=y+z+2,x²=y²+z²+2yz+200 求 x+y+z=? 方程组xy/x+y=2 yz/y+z=4 zx/x+z=5 求x,y,z (2X+Z-Y)/(X^2-XY+XZ-YZ)-(Y-Z)/(X^2-XY-XZ+YZ) 求Y(Z)=Z(Z+2)/(3Z-7)(Z+1)的z反变换 x^2(y+z)^2-2xy(x-z)(y+z)+y^2(x-z)^2 =[x(y+z)-y(x-z)]^2 =(xz+yz)^2 =z^2(x+y)^2 若|x-3|+|y+2|+|2z+1|=0,求(xy-yz)(y-x+z) 已知:x+y+z=a,xy+yz+zx=b,则x^2+y^2+z^2等于多少. 若z=z(x,y)由x^2+y^2+z^2=ye^yz确定 求dz 设z=z(x,y)由yz+x^2+z=0确定,求dz(x,y) x+y+z=2,xy+xz+yz=a,xy+z=a,求a的取值范围 分式题:xy=x+y,yz=2(y+z),zx=3(z+x),求xyz/(xy+yz+xz)xy=x+y,yz=2(y+z),zx=3(z+x),求xyz/(xy+yz+xz) f(x,y,z)=yz+xz使得,y^2+z^2=1,yz=3,求f最大值 求证不等式 xyz[yz(y+z)+zx(z+x)+xy(x+y)]>=2(xy+yz+xz)^2 若|x-3|+|y+z|+|2z+1|=0,求XY-YZ的值