Đặt \(2x+1=a\) (cho nó gọn)
\(PT\Leftrightarrow\left(6a-1\right)^2\left(3a-1\right)2a=2\)
\(\Leftrightarrow108a^4-72a^3+15a^2-x-1=0\)
\(\Leftrightarrow\left(108a^4-54a^3\right)+\left(-18a^3+9a^2\right)+\left(6a^2-3a\right)+\left(2a-1\right)=0\)
\(\Leftrightarrow\left(2a-1\right)\left(54a^3-9a^2+3a+1\right)=0\)
\(\Leftrightarrow\left(2a-1\right)\left[\left(54a^3+9a^2\right)+\left(-18a^2-3a\right)+\left(6a+1\right)\right]=0\)
\(\Leftrightarrow\left(2a-1\right)\left(6a+1\right)\left(9a^2-3a+1\right)=0\)
Dễ thấy \(9a^2-3a+1>0\)
\(\Rightarrow\left[{}\begin{matrix}a=\dfrac{1}{2}\\a=-\dfrac{1}{6}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x+1=\dfrac{1}{2}\\2x+1=-\dfrac{1}{6}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=-\dfrac{7}{12}\end{matrix}\right.\)
