\(d,-7x+1=-6x^2\)
\(\Rightarrow6x^2-7x+1=0\)
\(\Rightarrow6x^2-6x-x+1=0\)
\(\Rightarrow6x\left(x-1\right)-\left(x-1\right)=0\)
\(\Rightarrow\left(6x-1\right)\left(x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}6x-1=0\\x-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=1\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{1}{6};1\right\}\)