Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\left\{ \begin{array}{l}
x + y + xy = 19\\
y + z + yz = 11\\
z + x + zx = 14
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left( {x + xy} \right) + \left( {y + 1} \right) = 20\\
\left( {y + yz} \right) + \left( {z + 1} \right) = 12\\
\left( {z + zx} \right) + \left( {x + 1} \right) = 15
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x\left( {1 + y} \right) + \left( {y + 1} \right) = 20\\
y\left( {1 + z} \right) + \left( {z + 1} \right) = 12\\
z\left( {1 + x} \right) + \left( {x + 1} \right) = 15
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left( {x + 1} \right)\left( {y + 1} \right) = 20\\
\left( {y + 1} \right)\left( {z + 1} \right) = 12\\
\left( {z + 1} \right)\left( {x + 1} \right) = 15
\end{array} \right.\\
\Rightarrow \left[ {\left( {x + 1} \right)\left( {y + 1} \right)} \right].\left[ {\left( {y + 1} \right)\left( {z + 1} \right)} \right].\left[ {\left( {z + 1} \right)\left( {x + 1} \right)} \right] = 20.12.15\\
\Leftrightarrow {\left[ {\left( {x + 1} \right)\left( {y + 1} \right)\left( {z + 1} \right)} \right]^2} = 3600\\
\Leftrightarrow \left[ \begin{array}{l}
\left( {x + 1} \right)\left( {y + 1} \right)\left( {z + 1} \right) = 60\\
\left( {x + 1} \right)\left( {y + 1} \right)\left( {z + 1} \right) = - 60
\end{array} \right.\\
TH1:\,\,\left( {x + 1} \right)\left( {y + 1} \right)\left( {z + 1} \right) = 60\\
\Rightarrow \left\{ \begin{array}{l}
z + 1 = \dfrac{{60}}{{20}} = 3\\
y + 1 = \dfrac{{60}}{{15}} = 4\\
x + 1 = \dfrac{{60}}{{12}} = 5
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 4\\
y = 3\\
z = 2
\end{array} \right.\\
TH2:\,\,\,\left( {x + 1} \right)\left( {y + 1} \right)\left( {z + 1} \right) = - 60\\
\Rightarrow \left\{ \begin{array}{l}
z + 1 = \dfrac{{ - 60}}{{20}} = - 3\\
y + 1 = \dfrac{{ - 60}}{{15}} = - 4\\
x + 1 = \dfrac{{ - 60}}{{12}} = - 5
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = - 6\\
y = - 5\\
z = - 4
\end{array} \right.
\end{array}\)
