Question

\(chox,y,z\ne0;x\ne y;\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}CM:x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)

Answer 1

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Answer 2

\(\frac{x^2-yz}{x\left(1-yz\right)}=\frac{y^2-xz}{y\left(1-xz\right)}\)

\(\Leftrightarrow\left(x^2-yz\right)\left(y-xyz\right)=\left(y^2-xz\right)\left(x-xyz\right)\)

\(\Leftrightarrow x^2y-x^3yz-y^2z+xy^2z^2-xy^2+xy^3z+x^2z-x^2yz^2=0\)

\(\Leftrightarrow xy\left(x-y\right)-xyz\left(x^2-y^2\right)+z\left(x^2-y^2\right)-xyz^2\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left[xy-xyz\left(x+y\right)+z\left(x+y\right)-xyz^2\right]=0\)

\(\Leftrightarrow xy-xyz\left(x+y\right)+z\left(x+y\right)-xyz^2=0\left(x\ne y\Rightarrow x-y\ne0\right)\)

\(\Leftrightarrow xy+yz+xz=xyz\left(x+y\right)+xyz^2\)

\(\Leftrightarrow\frac{ay+yz+xz}{xyz}=\frac{xyz\left(x+y\right)+xyz^2}{xyz}\left(xyz\ne0\right)\)

\(\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=x+y+z\)