ĐKXĐ: x<>0; y<>0
\(\left\{{}\begin{matrix}\dfrac{3}{5x}+\dfrac{1}{y}=\dfrac{1}{10}\\\dfrac{3}{4x}+\dfrac{3}{4y}=\dfrac{1}{12}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{9}{20x}+\dfrac{3}{4y}=\dfrac{1}{10}\cdot\dfrac{3}{4}=\dfrac{3}{40}\\\dfrac{3}{4x}+\dfrac{3}{4y}=\dfrac{1}{12}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{9}{20x}+\dfrac{3}{4y}-\dfrac{3}{4x}-\dfrac{3}{4y}=\dfrac{3}{40}-\dfrac{1}{12}=\dfrac{9}{120}-\dfrac{10}{120}=-\dfrac{1}{120}\\\dfrac{3}{5x}+\dfrac{1}{y}=\dfrac{1}{10}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{9}{20x}-\dfrac{15}{20x}=-\dfrac{1}{120}\\\dfrac{3}{5x}+\dfrac{1}{y}=\dfrac{1}{10}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{6}{20x}=-\dfrac{1}{120}\\\dfrac{1}{y}=\dfrac{1}{10}-\dfrac{3}{5x}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}20x=6\cdot120=720\\\dfrac{1}{y}=\dfrac{1}{10}-\dfrac{3}{5x}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=36\\\dfrac{1}{y}=\dfrac{1}{10}-\dfrac{3}{5\cdot36}=\dfrac{1}{10}-\dfrac{1}{60}=\dfrac{5}{60}=\dfrac{1}{12}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=36\\y=12\end{matrix}\right.\left(nhận\right)\)
ĐKXĐ: \(x;y\ne0\)
Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x}=u\\\dfrac{1}{y}=v\end{matrix}\right.\) ta được:
\(\left\{{}\begin{matrix}\dfrac{3}{5}u+v=\dfrac{1}{10}\\\dfrac{3}{4}u+\dfrac{3}{4}v=\dfrac{1}{12}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{5}u+v=\dfrac{1}{10}\\u+v=\dfrac{1}{9}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{5}u=\dfrac{1}{90}\\v=\dfrac{1}{9}-u\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}u=\dfrac{1}{36}\\v=\dfrac{1}{9}-\dfrac{1}{36}=\dfrac{1}{12}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{1}{36}\\\dfrac{1}{y}=\dfrac{1}{12}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=36\\y=12\end{matrix}\right.\)
