ĐKXĐ:...
Bình phương 2 vế pt đầu và rút gọn ta được:
\(\sqrt{\left(2x-y-1\right)\left(3y+1\right)}=\sqrt{x\left(x+2y\right)}\)
\(\Leftrightarrow\left(2x-y-1\right)\left(3y+1\right)=x\left(x+2y\right)\)
\(\Leftrightarrow6xy-3y^2-3y+2x-y-1=x^2+2xy\)
\(\Leftrightarrow x^2-4xy+3y^2+4y-2x+1=0\)
\(\Leftrightarrow\left(x-2y-1\right)^2-y^2=0\)
\(\Leftrightarrow\left(x-3y-1\right)\left(x-y-1\right)=0\Rightarrow\left[{}\begin{matrix}x=3y+1\\x=y+1\end{matrix}\right.\)
Thay vào pt dưới: \(\left(x-1\right)^2\left(x+2\right)=2y^3-3y\)
\(\Leftrightarrow\left[{}\begin{matrix}9y^2\left(3y+3\right)=2y^3-y^2\\y^2\left(y+3\right)=2y^3-y^2\end{matrix}\right.\) \(\Leftrightarrow...\)