Đáp án:
$\begin{array}{l}
1)7x\left( {x - 4} \right) - x + 4 = 0\\
\Leftrightarrow \left( {x - 4} \right)\left( {7x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 4\\
x = \dfrac{1}{7}
\end{array} \right.\\
Vậy\,x = \dfrac{1}{7};x = 4\\
2){x^3} - 3{x^2} + x - 3 = 0\\
\Leftrightarrow {x^2}\left( {x - 3} \right) + x - 3 = 0\\
\Leftrightarrow \left( {x - 3} \right)\left( {{x^2} + 1} \right) = 0\\
\Leftrightarrow x - 3 = 0\\
\Leftrightarrow x = 3\\
Vậy\,x = 3\\
3){x^3} - 2{x^2} - 9x + 18 = 0\\
\Leftrightarrow {x^2}\left( {x - 2} \right) - 9\left( {x - 2} \right) = 0\\
\Leftrightarrow \left( {x - 2} \right)\left( {{x^2} - 9} \right) = 0\\
\Leftrightarrow \left( {x - 2} \right)\left( {x - 3} \right)\left( {x + 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = 3\\
x = - 3
\end{array} \right.\\
Vậy\,x = 2;x = 3;x = - 3\\
4)3x\left( {x - 1} \right) + x - 1 = 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {3x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = - \dfrac{1}{3}
\end{array} \right.\\
Vậy\,x = - \dfrac{1}{3};x = 1\\
5)2\left( {x + 3} \right) - {x^2} - 3x = 0\\
\Leftrightarrow 2\left( {x + 3} \right) - x\left( {x + 3} \right) = 0\\
\Leftrightarrow \left( {x + 3} \right)\left( {2 - x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 3\\
x = 2
\end{array} \right.\\
Vậy\,x = - 3;x = 2\\
6)6{x^2} - 15x - \left( {2x - 5} \right)\left( {x + 5} \right) = 0\\
\Leftrightarrow 3x\left( {2x - 5} \right) - \left( {2x - 5} \right)\left( {x + 5} \right) = 0\\
\Leftrightarrow \left( {2x - 5} \right)\left( {3x - x - 5} \right) = 0\\
\Leftrightarrow \left( {2x - 5} \right)\left( {2x - 5} \right) = 0\\
\Leftrightarrow 2x - 5 = 0\\
\Leftrightarrow x = \dfrac{5}{2}\\
Vậy\,x = \dfrac{5}{2}
\end{array}$
