Từ \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)
\(\Rightarrow\) \(\frac{yz+xz+xy}{xyz}=\frac{1}{x+y+z}\)
\(\Leftrightarrow\) \(yz\left(x+y+z\right)+xz\left(x+y+z\right)+xy\left(x+y+z\right)=xyz\)
\(\Leftrightarrow\) \(xyz+y^2z+yz^2+x^2z+xyz+xz^2+x^2y+xy^2+xyz-xyz=0\)
\(\Leftrightarrow\) \(2xyz+y^2z+yz^2+x^2z+xz^2+x^2y+xy^2=0\)
\(\Leftrightarrow\) \(x^2\left(y+z\right)+x\left(y^2+2yz+z^2\right)+yz\left(y+z\right)=0\)
\(\Leftrightarrow\) \(\left(y+z\right)\left(x^2+xy+xz+yz\right)=0\)
\(\Leftrightarrow\) \(\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)
\(\Leftrightarrow\) \(x=-y\) hoặc \(y=-z\) hoặc \(z=-x\)
Vậy, trong ba số x, y, z có hai số đối nhau
