\(\left(6x+7\right)^2\left(3x+4\right)\left(x+1\right)=6\)

\(\Rightarrow\left(6x+7\right)^2.2.\left(3x+4\right).6.\left(x+1\right)=72\)

\(\Rightarrow\left(6x+7\right)^2\left(6x+8\right)\left(6x+6\right)=72\)

\(\Rightarrow\left(6x+7\right)^2\left(6x+7+1\right)\left(6x+7-1\right)=72\)

\(\Rightarrow\left(6x+7\right)^2\left[\left(6x+7\right)^2-1\right]=72\)

\(\Rightarrow\left(6x+7\right)^4-\left(6x+7\right)^2=72\)

\(\Rightarrow\left(6x+7\right)^4-9\left(6x+7\right)^2+8\left(6x+7\right)^2-72=0\)

\(\Rightarrow\left(6x+7\right)^2\left[\left(6x+7\right)^2-9\right]+8\left[\left(6x+7\right)^2-9\right]=0\)

\(\Rightarrow\left[\left(6x+7\right)^2+8\right]\left[\left(6x+7\right)^2-9\right]=0\)

\(\Rightarrow\left(6x+7\right)^2-9=0\) Vì \(\left(6x+7\right)^2+8>0\) với mọi \(x\)

\(\Rightarrow\left(6x+7\right)^2=9\Rightarrow6x+7=3\) hoặc \(-3\)

\(\Rightarrow\left[\begin{matrix}6x+7=3\Rightarrow x=\frac{-2}{3}\\6x+7=-3\Rightarrow x=\frac{-5}{3}\end{matrix}\right.\)

\(\Rightarrow x=\frac{-2}{3};\frac{-5}{3}\)

a)Đk:\(x\ge\frac{1}{2}\)

\(pt\Leftrightarrow4x^2-12x+4+4\sqrt{2x-1}=0\)

\(\Leftrightarrow\left(2x-1\right)^2-4\left(2x-1\right)-1+4\sqrt{2x-1}=0\)

Đặt \(t=\sqrt{2x-1}>0\Rightarrow\hept{\begin{cases}t^2=2x-1\\t^4=\left(2x-1\right)^2\end{cases}}\)

\(t^4-4t^2+4t-1=0\)

\(\Leftrightarrow\left(t-1\right)^2\left(t^2+2t-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}t-1=0\\t^2+2t-1=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}t=1\\t=\sqrt{2}-1\end{cases}\left(t>0\right)}\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x=2-\sqrt{2}\end{cases}}\) là nghiệm thỏa pt

a) \(3x^3+6x^2-4x=0\) \(\Leftrightarrow\) \(x\left(3x^2+6x-4\right)=0\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=0\\3x^2+6x-4=0\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x=0\\\left\{{}\begin{matrix}x=\dfrac{-3+\sqrt{21}}{3}\\x=\dfrac{-3-\sqrt{21}}{3}\end{matrix}\right.\end{matrix}\right.\)

vậy phương trình có 2 nghiệm \(x=0;x=\dfrac{-3+\sqrt{21}}{3};x=\dfrac{-3-\sqrt{21}}{3}\)