Đáp án:
$\begin{array}{l}
1){\left( {x - 2} \right)^2} - \left( {x + 1} \right)\left( {x + 2} \right) = - 7\\
\Leftrightarrow {x^2} - 4x + 4 - \left( {{x^2} + 3x + 2} \right) = - 7\\
\Leftrightarrow {x^2} - 4x + 4 - {x^2} - 3x - 2 = - 7\\
\Leftrightarrow - 7x = - 9\\
\Leftrightarrow x = \dfrac{9}{7}\\
Vậy\,x = \dfrac{9}{7}\\
2)4{x^2} - 4x + 1 = {\left( {x - 2} \right)^2}\\
\Leftrightarrow {\left( {2x - 1} \right)^2} = {\left( {x - 2} \right)^2}\\
\Leftrightarrow \left[ \begin{array}{l}
2x - 1 = x - 2\\
2x - 1 = - x + 2
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 1\\
x = 1
\end{array} \right.\\
Vậy\,x = - 1;x = 1\\
3){x^2} - 5x = 6\\
\Leftrightarrow {x^2} - 5x - 6 = 0\\
\Leftrightarrow \left( {x - 6} \right)\left( {x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 6\\
x = - 1
\end{array} \right.\\
Vậy\,x = - 1;x = 6\\
4)3{x^2}\left( {x - 7} \right) + 2\left( {7 - x} \right) = 0\\
\Leftrightarrow \left( {x - 7} \right)\left( {3{x^2} - 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 7\\
{x^2} = \dfrac{2}{3}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 7\\
x = \pm \dfrac{{\sqrt 6 }}{3}
\end{array} \right.\\
Vậy\,x = 7;x = \pm \dfrac{{\sqrt 6 }}{3}\\
5)3x\left( {4 - x} \right) + 3{x^2} = 36\\
\Leftrightarrow 12x - 3{x^2} + 3{x^2} - 36 = 0\\
\Leftrightarrow 12x = 36\\
\Leftrightarrow x = 3\\
Vậy\,x = 3
\end{array}$
