Đáp án:
$\begin{array}{l}
1){\left( {x - 2} \right)^2} - \left( {x + 1} \right)\left( {x + 3} \right) = - 7\\
\Leftrightarrow {x^2} - 4x + 4 - \left( {{x^2} + 4x + 3} \right) = - 7\\
\Leftrightarrow 7 = - 7\left( {ktm} \right)\\
Vậypt\\
2)\left( {3x + 5} \right)\left( {4 - 3x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
3x + 5 = 0\\
4 - 3x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{ - 5}}{3}\\
x = \dfrac{4}{3}
\end{array} \right.\\
Vậy\,x = \dfrac{{ - 5}}{3};x = \dfrac{4}{3}\\
3)3x\left( {x - 7} \right) - 2\left( {x - 7} \right) = 0\\
\Leftrightarrow \left( {x - 7} \right)\left( {3x - 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 7 = 0\\
3x - 2 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 7\\
x = \dfrac{2}{3}
\end{array} \right.\\
Vậy\,x = \dfrac{2}{3};x = 7\\
4)8{x^2} - 8x + 2 = 0\\
\Leftrightarrow 4{x^2} - 4x + 1 = 0\\
\Leftrightarrow {\left( {2x - 1} \right)^2} = 0\\
\Leftrightarrow x = \dfrac{1}{2}\\
Vậy\,x = \dfrac{1}{2}\\
5)7{x^2} - 28 = 0\\
\Leftrightarrow {x^2} = 4\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x = - 2
\end{array} \right.\\
Vậy\,x = 2;x = - 2\\
6)3{x^3} - 27x = 0\\
\Leftrightarrow 3x\left( {{x^2} - 9} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
{x^2} = 9
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 3;x = - 3
\end{array} \right.\\
Vậy\,x = 0;x = 3;x = - 3\\
7)\dfrac{2}{3}x\left( {{x^2} - 4} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
{x^2} = 4
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 2;x = - 2
\end{array} \right.\\
Vậy\,x = 0;x = 2;x = - 2\\
8)9{x^3} + 9 = 0\\
\Leftrightarrow {x^3} = - 1\\
\Leftrightarrow x = - 1\\
Vậy\,x = - 1
\end{array}$
