Giải thích các bước giải:
Đặt \(\dfrac{{xy + xz}}{2} = \dfrac{{yx + yz}}{3} = \dfrac{{zx + zy}}{4} = t\) ta có:
\(\begin{array}{l}
\dfrac{{xy + xz}}{2} = \dfrac{{yx + yz}}{3} = \dfrac{{zx + zy}}{4} = t\\
\Rightarrow \left\{ \begin{array}{l}
xy + xz = 2t\\
yx + yz = 3t\\
zx + zy = 4t
\end{array} \right.\\
\Rightarrow \left( {xy + xz} \right) + \left( {yx + yz} \right) + \left( {zx + zy} \right) = 2t + 3t + 4t\\
\Leftrightarrow 2.\left( {xy + yz + zx} \right) = 9t\\
\Rightarrow xy + yz + zx = \dfrac{{9t}}{2}\\
\Rightarrow \left\{ \begin{array}{l}
xy = \left( {xy + yz + zx} \right) - \left( {zx + zy} \right)\\
yz = \left( {xy + yz + zx} \right) - \left( {xy + zx} \right)\\
zx = \left( {xy + yz + zx} \right) - \left( {xy + yz} \right)
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
xy = \dfrac{{9t}}{2} - 4t = \dfrac{t}{2}\\
yz = \dfrac{{9t}}{2} - 2t = \dfrac{{5t}}{2}\\
zx = \dfrac{{9t}}{2} - 3t = \dfrac{{3t}}{2}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
\dfrac{x}{z} = \dfrac{{xy}}{{yz}} = \dfrac{1}{5}\\
\dfrac{x}{y} = \dfrac{{xz}}{{yz}} = \dfrac{3}{5}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{1}{5}z\\
x = \dfrac{3}{5}y
\end{array} \right.\\
\Rightarrow x = \dfrac{{3y}}{5} = \dfrac{z}{5}\\
\Leftrightarrow \dfrac{x}{3} = \dfrac{y}{5} = \dfrac{z}{{15}}
\end{array}\)
