\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{5}{24}\left(1\right)\\\dfrac{9}{x}+\dfrac{6}{5}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)=1\left(2\right)\end{matrix}\right.\left(x\ne0,y\ne0\right)\)
Thay (1) vào (2) ta có: \(\dfrac{9}{x}+\dfrac{6}{5}\cdot\dfrac{5}{24}=1\Leftrightarrow\dfrac{9}{x}+\dfrac{1}{4}=1\Leftrightarrow\dfrac{9}{x}=\dfrac{3}{4}\)\(\Leftrightarrow x=12\left(TM\right)\)
Thay \(x=12\)vào (1) ta có: \(\dfrac{1}{12}+\dfrac{1}{y}=\dfrac{5}{24}\Leftrightarrow\dfrac{1}{y}=\dfrac{1}{8}\Leftrightarrow y=8\left(TM\right)\)
Vậy HPT có nghiệm (12;8)