` 9x^2 - (6x+2)(x-5) =1`
`\to 9x^2 -(6x^2 - 30x +2x -10) = 1`
`\to 9x^2 - (6x^2 - 28x -10) =1`
`\to 9x^2 - 6x^2 + 28x + 10 - 1 = 0`
`\to 3x^2 +28x + 9 = 0`
`\to 3x^2 + 27x + x + 9 = 0`
`\to 3x(x+9) + (x+9)=0`
`\to (3x+1)(x+9) = 0`
`\to` \(\left[ \begin{array}{l}3x+1=0\\\\x+9=0\end{array} \right.\) `\to` \(\left[ \begin{array}{l}3x=-1\\\\x=-9\end{array} \right.\) `\to` \(\left[ \begin{array}{l}x= \dfrac{-1}{3}\\\\x=-9\end{array} \right.\)
Vậy ` x \ in { -1/3 ; -9}`