a.
ĐKXĐ: \(\left\{{}\begin{matrix}x\ne2\\y\ne\dfrac{1}{2}\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x-2}=u\\\dfrac{1}{2y-1}=v\end{matrix}\right.\) ta được:
\(\left\{{}\begin{matrix}u+v=2\\2u-3v=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3u+3v=6\\2u-3v=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5u=7\\2u-3v=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}u=\dfrac{7}{5}\\v=\dfrac{3}{5}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x-2}=\dfrac{7}{5}\\\dfrac{1}{2y-1}=\dfrac{3}{5}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x-2=\dfrac{5}{7}\\2y-1=\dfrac{5}{3}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{19}{7}\\y=\dfrac{4}{3}\end{matrix}\right.\)
b.
ĐKXĐ: \(\left\{{}\begin{matrix}2x+y\ne0\\x-2y\ne0\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}\dfrac{1}{2x+y}=u\\\dfrac{1}{x-2y}=v\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}u+v=\dfrac{5}{8}\\u-v=-\dfrac{3}{8}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u=\dfrac{1}{8}\\v=\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2x+y}=\dfrac{1}{8}\\\dfrac{1}{x-2y}=\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x+y=8\\x-2y=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{18}{5}\\y=\dfrac{4}{5}\end{matrix}\right.\)
