Đáp án:
\(\left[ \begin{array}{l}
y = 10\\
y = \dfrac{3}{2}
\end{array} \right. \to \left[ \begin{array}{l}
x = 6\\
x = - \dfrac{5}{2}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ne 0;y \ne 0\\
\left\{ \begin{array}{l}
\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{4}{{15}}\\
x = y - 4
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = y - 4\\
\dfrac{1}{{y - 4}} + \dfrac{1}{y} = \dfrac{4}{{15}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = y - 4\\
\dfrac{{15y + 15y - 15.4 - 4y\left( {y - 4} \right)}}{{15y\left( {y - 4} \right)}} = 0\left( {DK:y \ne \left\{ {0;4} \right\}} \right)
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = y - 4\\
- 4{y^2} + 46y - 60 = 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = y - 4\\
\left[ \begin{array}{l}
y = 10\\
y = \dfrac{3}{2}
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 6\\
x = - \dfrac{5}{2}
\end{array} \right.
\end{array}\)
