ĐKXĐ: \(x\ge\dfrac{-1}{3};x\ne0\)
\(3x-\left(1-\dfrac{x-1}{4x}\right)=\sqrt{3x+1}\)
\(\Leftrightarrow-3x+\dfrac{3x+1}{4x}+\sqrt{3x+1}=0\)
\(\Leftrightarrow\dfrac{-3}{4}+\dfrac{3x+1}{\left(4x\right)^2}+\dfrac{\sqrt{3x+1}}{4x}=0\)
Đặt \(\dfrac{\sqrt{3x+1}}{4x}=a\Rightarrow a^2+a-\dfrac{3}{4}=0\Rightarrow\left[{}\begin{matrix}a=\dfrac{-3}{2}\\a=\dfrac{1}{2}\end{matrix}\right.\)
TH1: \(a=\dfrac{-3}{2}\Rightarrow\dfrac{\sqrt{3x+1}}{4x}=-\dfrac{3}{2}\Leftrightarrow\sqrt{3x+1}=-6x\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le0\\3x+1=36x^2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\le0\\36x^2-3x-1=0\end{matrix}\right.\) \(\Rightarrow x=\dfrac{1-\sqrt{17}}{24}\)
TH2: \(a=\dfrac{1}{2}\Rightarrow\dfrac{\sqrt{3x+1}}{4x}=\dfrac{1}{2}\Leftrightarrow\sqrt{3x+1}=2x\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\3x+1=4x^2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\4x^2-3x-1=0\end{matrix}\right.\) \(\Rightarrow x=1\)
