ĐKXĐ:\(x\ge-\dfrac{1}{2}\)
\(6x^2+2x+1=3x\sqrt{6x+3}\\ \Leftrightarrow\left(6x^2+2x+1\right)^2=\left(3x\sqrt{6x+3}\right)^2\\ \Leftrightarrow36x^4+24x^3+16x^2+4x+1=9x^2\left(6x+3\right)\\ \Leftrightarrow36x^4+24x^3+16x^2+4x+1=54x^3+27x^2\\ \Leftrightarrow36x^4-30x^3-11x^2+4x+1=0\\ \Leftrightarrow...\)