\(\sqrt{2x+1}< 2\Leftrightarrow\left[{}\begin{matrix}2x+1< 4\\2x+1>0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x< 3\\2x>-1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x< \dfrac{3}{2}\\x>\dfrac{-1}{2}\end{matrix}\right.\)
\(\Rightarrow\dfrac{-1}{2}< x< \dfrac{3}{2}\) vậy \(\dfrac{-1}{2}< x< \dfrac{3}{2}\)
\(\sqrt{2x+1}< 2=>\sqrt{2x+1}^2< 4\)
2x+1<4 => 2x<3 => x<3/2
\(\sqrt{2x+1}< 2\Leftrightarrow2x+1< 4\Leftrightarrow2x< 3\Leftrightarrow x< \dfrac{3}{2}\)
