ĐKXĐ: x<>0; y<>0
\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=3\\\dfrac{2018}{x}-\dfrac{2017}{y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=3-\dfrac{1}{y}\\2018\left(3-\dfrac{1}{y}\right)-\dfrac{2017}{y}=1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{1}{x}=3-\dfrac{1}{y}\\6054-\dfrac{2018}{y}-\dfrac{2017}{y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=3-\dfrac{1}{y}\\\dfrac{4035}{y}=6054-1=6053\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=\dfrac{4035}{6053}\\\dfrac{1}{x}=3-\dfrac{6053}{4035}=\dfrac{6052}{4035}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{4035}{6053}\\x=\dfrac{4035}{6052}\end{matrix}\right.\left(nhận\right)\)
