Đáp án:

$\begin{array}{l}
7)4\left( {2x - 1} \right) - {\left( {1 - 2x} \right)^2} = 0\\
 \Leftrightarrow 4\left( {2x - 1} \right) - {\left( {2x - 1} \right)^2} = 0\\
 \Leftrightarrow \left( {2x - 1} \right)\left( {4 - 2x + 1} \right) = 0\\
 \Leftrightarrow \left( {2x - 1} \right)\left( {5 - 2x} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
2x = 1\\
2x = 5
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x = \dfrac{5}{2}
\end{array} \right.\\
\text{Vậy}\,x = \dfrac{1}{2};x = \dfrac{5}{2}\\
8){\left( {3x - 2} \right)^2} = {\left( {4 - x} \right)^2}\\
 \Leftrightarrow \left[ \begin{array}{l}
3x - 2 = 4 - x\\
3x - 2 = x - 4
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
4x = 6\\
2x =  - 2
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{3}{2}\\
x =  - 1
\end{array} \right.\\
\text{Vậy}\,x = \dfrac{3}{2};x =  - 1\\
9)x - 2{x^2} + {x^3} = 0\\
 \Leftrightarrow x\left( {{x^2} - 2x + 1} \right) = 0\\
 \Leftrightarrow x{\left( {x - 1} \right)^2} = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 1
\end{array} \right.\\
\text{Vậy}\,x = 0;x = 1\\
10)3x - {x^2} - 2\left( {x - 3} \right) = 0\\
 \Leftrightarrow  - x\left( {x - 3} \right) - 2\left( {x - 3} \right) = 0\\
 \Leftrightarrow \left( {x - 3} \right)\left( { - x - 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x =  - 2
\end{array} \right.\\
\text{Vậy}\,x = 3;x =  - 2\\
11)4{x^2} - 25 = \left( {5 - 2x} \right)\left( {2x + 7} \right)\\
 \Leftrightarrow \left( {2x - 5} \right)\left( {2x + 5} \right) + \left( {2x - 5} \right)\left( {2x + 7} \right) = 0\\
 \Leftrightarrow \left( {2x - 5} \right)\left( {2x + 5 + 2x + 7} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{5}{2}\\
4x + 12 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{5}{2}\\
x =  - 3
\end{array} \right.\\
\text{Vậy}\,x =  - 3;x = \dfrac{5}{2}\\
12){x^3} + 27 + \left( {x + 3} \right)\left( {x - 9} \right) = 0\\
 \Leftrightarrow \left( {x + 3} \right)\left( {{x^2} - 3x + 9} \right) + \left( {x + 3} \right)\left( {x - 9} \right) = 0\\
 \Leftrightarrow \left( {x + 3} \right)\left( {{x^2} - 3x + 9 + x - 9} \right) = 0\\
 \Leftrightarrow \left( {x + 3} \right)\left( {{x^2} - 2x} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - 3\\
x = 0\\
x = 2
\end{array} \right.\\
\text{Vậy}\,x =  - 3;x = 0;x = 2\\
13)4{\left( {2x + 7} \right)^2} - 9{\left( {x + 3} \right)^2} = 0\\
 \Leftrightarrow {\left( {4x + 14} \right)^2} - {\left( {3x + 9} \right)^2} = 0\\
 \Leftrightarrow \left( {4x + 14 - 3x - 9} \right)\left( {4x + 14 + 3x + 9} \right) = 0\\
 \Leftrightarrow \left( {x + 5} \right)\left( {7x + 23} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - 5\\
x =  - \dfrac{{23}}{7}
\end{array} \right.\\
\text{Vậy}\,x =  - 5;x = \dfrac{{ - 23}}{7}\\
14){\left( {5{x^2} + 3x - 2} \right)^2} = {\left( {4{x^2} - 3x - 2} \right)^2}\\
 \Leftrightarrow \left[ \begin{array}{l}
5{x^2} + 3x - 2 = 4{x^2} - 3x - 2\\
5{x^2} + 3x - 2 =  - 4{x^2} + 3x + 2
\end{array} \right.
\end{array}$

$\begin{array}{l}
 \Leftrightarrow \left[ \begin{array}{l}
{x^2} + 6x = 0\\
9{x^2} = 4
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x\left( {x + 6} \right) = 0\\
{x^2} = \dfrac{4}{9}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x =  - 6\\
x = \dfrac{2}{3}\\
x =  - \dfrac{2}{3}
\end{array} \right.\\
\text{Vậy}\,x \in \left\{ { - 6; - \dfrac{2}{3};0;\dfrac{2}{3}} \right\}
\end{array}$