ĐKXĐ: \(x>\dfrac{1}{2}\)
Đặt \(\dfrac{1}{\sqrt{2x-1}}=z>0\) ta được:
\(\left\{{}\begin{matrix}4z+2\left(y+1\right)=\dfrac{22}{3}\\z-3\left(y-2\right)=\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4z+2y=\dfrac{16}{3}\\z-3y=-\dfrac{17}{3}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}z=\dfrac{1}{3}\\y=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{\sqrt{2x-1}}=\dfrac{1}{3}\\y=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{2x-1}=3\\y=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2x-1=9\\y=2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=5\\y=2\end{matrix}\right.\)
