Đặt \(\left\{{}\begin{matrix}\sqrt{2x+1}=a\\\frac{y}{3y+2}=b\end{matrix}\right.\)
ĐKXĐ: ....
\(\Leftrightarrow\left\{{}\begin{matrix}a-b=1\\3a-5b=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\)
\(\sqrt{2x+1}=2\Rightarrow x=1.5\)
\(\frac{y}{3y+2}=1\Rightarrow y=-1\)