mình nhầm \(12x\sqrt{3x+1}+12x+\sqrt{3x+1}+2=0\)
nhé!
ĐKXĐ: \(x\ge-\frac{1}{3}\)
Đặt \(\sqrt{3x+1}=a\ge0\Rightarrow3x=a^2-1\)
\(\Leftrightarrow4\left(a^2-1\right)a+4\left(a^2-1\right)+a+2=0\)
\(\Leftrightarrow4a^3+4a^2-3a-2=0\)
\(\Leftrightarrow4a^3+2a^2-4a+2a^2+a-2=0\)
\(\Leftrightarrow2a\left(2a^2+a-2\right)+\left(2a^2+a-2\right)=0\)
\(\Leftrightarrow\left(2a+1\right)\left(2a^2+a-2\right)=0\)
\(\Leftrightarrow2a^2+a-2=0\) (do \(a\ge0\Rightarrow2a+1>0\))
\(\Leftrightarrow\left[{}\begin{matrix}a=\frac{-1+\sqrt{17}}{4}\\a=\frac{-1-\sqrt{17}}{4}< 0\left(l\right)\end{matrix}\right.\) \(\Rightarrow\sqrt{3x+1}=\frac{-1+\sqrt{17}}{2}\)
\(\Rightarrow3x+1=\frac{9-\sqrt{17}}{8}\Rightarrow x=\frac{1-\sqrt{17}}{24}\)
