\((x-1)(2x-1) = 9-x\)

=> \(2x^2 - 3x + 1 = 9- x\)

=> \(2x^2 - 2x = 8\)

=> \(2x^2 - 2x - 8 = 0\)

=> \(2(x^2 - x - 4)= 0\)

=> \(2\left(x^2-x+\dfrac{1}{4}\right)-\dfrac{17}{2}=0\)

=> \(2\left(x-\dfrac{1}{2}\right)^2=\dfrac{17}{2}\)

=> \(\left(x-\dfrac{1}{2}\right)^2=\dfrac{17}{4}\)

=> \(\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{\sqrt{17}}{2}\\x-\dfrac{1}{2}=-\dfrac{\sqrt{17}}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{17}}{2}\\x=\dfrac{1-\sqrt{17}}{2}\end{matrix}\right.\)

\(\left(x-1\right)\left(2x-1\right)=9-x\)

\(\Leftrightarrow2x^3-3x+1=9-x\)

\(\Leftrightarrow2x^2-3x+1=9-x+x\)

\(\Leftrightarrow2x^2+2x+1=9\)

\(\Leftrightarrow2x^2-2x+1-9=9-9\)

\(\Leftrightarrow2x^2-2x-8=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\left(-2\right)+\sqrt{\left(-2\right)^2-4.2\left(-8\right)}}{2.2}\\x=\dfrac{-\left(2\right)+\sqrt{\left(-2\right)^2-4.\left(-8\right)}}{2.2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{17}}{2}\\\dfrac{1-\sqrt{17}}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{17}}{2}\\x=\dfrac{1-\sqrt{17}}{2}\end{matrix}\right.\)