Bài làm:
a) đk: \(x\ge-\frac{1}{2}\)
Ta có: \(\sqrt{2x+1}< 3\)
\(\Leftrightarrow2x+1< 9\)
\(\Leftrightarrow2x< 8\)
\(\Rightarrow x< 4\)
Vậy x < 4
b) đk: \(x\ge\frac{1}{3}\)
Ta có: \(\sqrt{3x-1}=\sqrt{x+2}\)
\(\Leftrightarrow\left|3x-1\right|=\left|x+2\right|\)
\(\Leftrightarrow\orbr{\begin{cases}3x-1=x+2\\3x-1=-x-2\end{cases}\Leftrightarrow}\orbr{\begin{cases}2x=3\\4x=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=-\frac{1}{4}\left(ktm\right)\end{cases}}\)
Vậy \(x=\frac{3}{2}\)
\(\sqrt{2x+1}< 3.\) ĐK: 2x+1 lớn hơn hoặc bằng 0 => x lớn hơn hoặc bằng -1/2
\(\Rightarrow\sqrt{2x+1}< \sqrt{9}\)
\(\Rightarrow2x+1< 9\)\(\Rightarrow x< 4\)
\(\Rightarrow-\frac{1}{2}\le x< 4\)
b/ \(\sqrt{3x-1}=\sqrt{x+2}\)( ĐK:x lớn hơn hoặc bằng 1/3)
\(\Rightarrow3x-1=x+2\)
\(\Rightarrow2x=3\Rightarrow x=\frac{3}{2}\left(tm\right)\)
