Đáp án:
\(\left[ \begin{array}{l}
y = 6\\
y = - 6
\end{array} \right. \to \left[ \begin{array}{l}
x = - 22\\
x = 22
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:y \ne 0\\
\left\{ \begin{array}{l}
\dfrac{{x + 2y}}{5} = \dfrac{{x + y}}{8}\\
\dfrac{{x + y}}{8} = - \dfrac{{12}}{y}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
8x + 16y = 5x + 5y\\
xy + {y^2} = - 96
\end{array} \right.\\
\to \left\{ \begin{array}{l}
3x + 11y = 0\\
xy + {y^2} = - 96
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = - \dfrac{{11y}}{3}\\
- \dfrac{{11y}}{3}.y + {y^2} = - 96
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = - \dfrac{{11y}}{3}\\
- \dfrac{8}{3}{y^2} = - 96
\end{array} \right.\\
\to \left\{ \begin{array}{l}
{y^2} = 36\\
x = - \dfrac{{11y}}{3}
\end{array} \right.\\
\to \left[ \begin{array}{l}
y = 6\\
y = - 6
\end{array} \right. \to \left[ \begin{array}{l}
x = - 22\\
x = 22
\end{array} \right.
\end{array}\)
