ĐK: \(x\ge\dfrac{-1}{3},x\ne0\)
pt\(\Leftrightarrow\)\(\dfrac{12x^2-4x+x-1}{4x}=\sqrt{3x+1}\)
\(\Rightarrow12x^2-3x-1=4x\sqrt{3x+1}\)
\(\Leftrightarrow\dfrac{3}{4}.16x^2-\left(3x+1\right)-\sqrt{16x^2\left(3x+1\right)}=0\)
Vì \(x\ne0\) nên chia cả 2 vế cho \(16x^2\), ta được:
\(\dfrac{3}{4}-\dfrac{3x+1}{16x^2}-\sqrt{\dfrac{3x+1}{16x^2}}=0\)
Đặt \(t=\sqrt{\dfrac{3x+1}{16x^2}}\left(t\ge0\right)\)
\(\Rightarrow t^2+t-\dfrac{3}{4}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{1}{2}\left(TM\right)\\t=\dfrac{-3}{2}\left(KTM\right)\end{matrix}\right.\)
Vậy \(\sqrt{\dfrac{3x+1}{16x^2}}=\dfrac{1}{2}\)
\(\Rightarrow4\left(3x+1\right)=16x^2\)
\(\Leftrightarrow16x^2-12x-4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{-1}{4}\end{matrix}\right.\)(TM)
Vậy pt có tập nghiệm là \(S=\left\{1;\dfrac{-1}{4}\right\}\)
