Giải hệ phương trình \(\left\{{}\begin{matrix}\dfrac{3}{5x}+\dfrac{1}{y}=\dfrac{1}{10}\\\dfrac{3}{4x}+\dfrac{3}{4y}=\dfrac{1}{12}\end{matrix}\right.\) , ta được nghiệm là
\(x=36,y=12\). \(x=\frac{1}{36},y=\frac{1}{12}\). \(x=2,y=5\). \(x=\dfrac{1}{24},y=\dfrac{1}{8}\), Hướng dẫn giải:\(\left\{{}\begin{matrix}\dfrac{3}{5x}+\dfrac{1}{y}=\dfrac{1}{10}\\\dfrac{3}{4x}+\dfrac{3}{4y}=\dfrac{1}{12}\end{matrix}\right.\) \(\left(ĐK:x,y\ne0\right)\)
Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x}=a\\\dfrac{1}{y}=b\end{matrix}\right.\left(a,b\ne0\right)\)
Phương trình đã cho trở thành: \(\left\{{}\begin{matrix}\dfrac{3}{5}a+b=\dfrac{1}{10}\\\dfrac{3}{4}a+\dfrac{3}{4}b=\dfrac{1}{12}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{9}{20}a+\dfrac{3}{4}b=\dfrac{3}{40}\\\dfrac{3}{4}a+\dfrac{3}{4}b=\dfrac{1}{12}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{5}a+b=\dfrac{1}{10}\\\dfrac{3}{10}a=\dfrac{1}{120}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{36}\\b=\dfrac{1}{12}\end{matrix}\right.\)
Suy ra \(\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{1}{36}\\\dfrac{1}{y}=\dfrac{1}{12}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=36\\y=12\end{matrix}\right.\left(tmđk\right)\)
Vậy hệ có nghiệm \(\left(x;y\right)=\left(36;12\right)\)