Đặt 1/x = a ; 1/y = b
\(\left\{{}\begin{matrix}a+b=\dfrac{1}{6}\\\dfrac{10}{3}a+10b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}10a+10b=\dfrac{5}{3}\\\dfrac{10}{3}a+10b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{20}{3}a=\dfrac{2}{3}\\b=\dfrac{1}{6}-a\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{10}\\b=\dfrac{1}{15}\end{matrix}\right.\)
Theo cách đặt x = 10 ; y = 15
ĐKXĐ:\(\left\{{}\begin{matrix}x\ne0\\y\ne0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{6}\\\dfrac{10}{3}.\dfrac{1}{x}+\dfrac{10}{y}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{6}\\\dfrac{10}{3x}+\dfrac{10}{y}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{6}\\\dfrac{1}{3x}+\dfrac{1}{y}=\dfrac{1}{10}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}-\dfrac{1}{3x}-\dfrac{1}{y}=\dfrac{1}{6}-\dfrac{1}{10}\\\dfrac{1}{3x}+\dfrac{1}{y}=\dfrac{1}{10}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{3x}=\dfrac{1}{15}\\\dfrac{1}{3x}+\dfrac{1}{y}=\dfrac{1}{10}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x=30\\\dfrac{1}{3x}+\dfrac{1}{y}=\dfrac{1}{10}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=10\\\dfrac{1}{3.10}+\dfrac{1}{y}=\dfrac{1}{10}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=10\\\dfrac{1}{30}+\dfrac{1}{y}=\dfrac{1}{10}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=10\\\dfrac{1}{y}=\dfrac{1}{15}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=15\end{matrix}\right.\)