ĐKXĐ:\(\left\{{}\begin{matrix}x\ne0\\y\ne0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y}=10\\\dfrac{1}{x}-\dfrac{2}{y}=15\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y}=10\\\dfrac{2}{x}-\dfrac{4}{y}=30\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y}-\dfrac{2}{x}+\dfrac{4}{y}=10-30\\\dfrac{2}{x}+\dfrac{3}{y}=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{7}{y}=-20\\\dfrac{2}{x}+\dfrac{3}{y}=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-7}{20}\\\dfrac{2}{x}+\dfrac{3}{-\dfrac{7}{20}}=10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-7}{20}\left(tm\right)\\x=\dfrac{7}{65}\left(tm\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{3}{y}=10.\\\dfrac{1}{x}-\dfrac{2}{y}=15.\end{matrix}\right.\) \(\left(x\ne0;y\ne0\right).\)
\(\Leftrightarrow\left\{{}\begin{matrix}2.\dfrac{1}{x}+3.\dfrac{1}{y}=10.\\\dfrac{1}{x}-2.\dfrac{1}{y}=15.\end{matrix}\right.\)
Đặt \(a=\dfrac{1}{x};b=\dfrac{1}{y}\left(a\ne0;b\ne0\right).\)
\(\Rightarrow\left\{{}\begin{matrix}2a+3b=10.\\a-2b=15.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2a+3b=10.\\2a-4b=30.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}7b=-20.\\a-2b=15.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=-\dfrac{20}{7}.\\a=\dfrac{65}{7}.\end{matrix}\right.\) \(\left(TM\right).\)
\(\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{20}{7}=\dfrac{1}{y}.\\\dfrac{65}{7}=\dfrac{1}{x}.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{7}{20}.\\x=\dfrac{7}{65}.\end{matrix}\right.\) \(\left(TM\right).\)
Vậy hệ phương trình có nghiệm duy nhất là \(\left(x;y\right)=\left(-\dfrac{7}{20};\dfrac{7}{65}\right).\)
