ĐKXĐ: \(xy\ne0\)
\(\left\{{}\begin{matrix}\dfrac{2}{x}-\dfrac{3}{y}=-2\\\dfrac{1}{x}+\dfrac{3}{y}=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x}=6\\\dfrac{1}{x}+\dfrac{3}{y}=8\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\\dfrac{1}{x}+\dfrac{3}{y}=8\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\2+\dfrac{3}{y}=8\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\\dfrac{3}{y}=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{1}{2}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{2}{x}-\dfrac{3}{y}=-2\\\dfrac{1}{x}+\dfrac{3}{y}=8\end{matrix}\right.\)
Lấy hệ pt 1 + hệ pt 2, ta được :
\(\Leftrightarrow\dfrac{3}{x}+0=6\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Thay \(x=\dfrac{1}{2}\) vào hệ pt 1 :
\(\dfrac{2}{\dfrac{1}{2}}-\dfrac{3}{y}=-2\)
\(\Leftrightarrow4-\dfrac{3}{y}=-2\)
\(\Leftrightarrow\dfrac{-3}{y}=-6\)
\(\Leftrightarrow y=\dfrac{1}{2}\)
Vậy hệ pt có nghiệm duy nhất \(\left(x;y\right)=\left(\dfrac{1}{2};\dfrac{1}{2}\right)\)
