\(2x\left(x+1\right)-\left(2x+1\right)\left(x-3\right)=6\)
\(2x^2+2x-2x^2-x+6x+3=0\)
\(7x=-3\) => \(x=-\frac{3}{7}\)
Ta có: \(\left(2x-1\right)^2-\left(3x+4\right)^2=0\)
\(\Leftrightarrow\left[\left(2x-1\right)+\left(3x+4\right)\right]\left[\left(2x-1\right)-\left(3x+4\right)\right]=0\)
\(\Leftrightarrow\left(2x-1+3x+4\right)\left(2x-1-3x-4\right)=0\)
\(\Leftrightarrow\left(5x+3\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+3=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-3\\x=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-3}{5}\\x=5\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{3}{5};5\right\}\)
