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
\(\left(x,y\right)=\left(36,12\right)\).\(\left(x,y\right)=\left(-13,39\right)\).\(\left(x,y\right)=\left(3,9\right)\).\(\left(x,y\right)=\left(11,33\right)\).Hướng dẫn giải:\(\left\{{}\begin{matrix}\dfrac{3}{5x}+\dfrac{1}{y}=\dfrac{1}{10}\\\dfrac{3}{4x}+\dfrac{3}{4y}=\dfrac{1}{12}\end{matrix}\right.\)
Đặt \(a=\dfrac{1}{x},b=\dfrac{1}{y}\) ta có hệ phương trình:
\(\left\{{}\begin{matrix}\dfrac{3}{5}a+b=\dfrac{1}{10}\\\dfrac{3}{4}a+\dfrac{3}{4}b=\dfrac{1}{12}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{5}a+b=\dfrac{1}{10}\\a+b=\dfrac{1}{9}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{36}\\b=\dfrac{1}{12}\end{matrix}\right.\).
Trở lại phép đặt ẩn phụ ta có:
\(\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{1}{36}\\\dfrac{1}{y}=\dfrac{1}{12}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=36\\y=12\end{matrix}\right.\)
