Đáp án:

$\begin{array}{l}
14)\\
6{x^2} - \left( {2x + 5} \right)\left( {3x - 2} \right) = 7\\
 \Leftrightarrow 6{x^2} - \left( {6{x^2} - 4x + 15x - 10} \right) = 7\\
 \Leftrightarrow 6{x^2} - 6{x^2} - 11x + 10 = 7\\
 \Leftrightarrow 11x = 3\\
 \Leftrightarrow x = \frac{3}{{11}}\\
Vậy\,x = \frac{3}{{11}}\\
15)\left( {2x - 1} \right)\left( {3x + 1} \right) + \left( {3x - 4} \right)\left( {3 - 2x} \right) = 5\\
 \Leftrightarrow 6{x^2} + 2x - 3x - 1 + 9x - 6{x^2} - 12 + 8x = 5\\
 \Leftrightarrow 16x = 18\\
 \Leftrightarrow x = \frac{9}{8}\\
Vậy\,x = \frac{9}{8}\\
16)\left( {4x + 7} \right)\left( {2 - 3x} \right) - \left( {6x + 2} \right)\left( {5 - 2x} \right) = 9\\
 \Leftrightarrow 8x - 12{x^2} + 14 - 21x\\
 - \left( {30x - 12{x^2} + 10 - 4x} \right) = 9\\
 \Leftrightarrow  - 12{x^2} - 13x + 14 + 12{x^2} - 26x - 10 = 9\\
 \Leftrightarrow 39x =  - 5\\
 \Leftrightarrow x = \frac{{ - 5}}{{39}}\\
Vậy\,x = \frac{{ - 5}}{{39}}\\
17)x\left( {x + \frac{1}{3}} \right) - \left( {\frac{1}{2}x + 4} \right)\left( {2x - 5} \right) = \frac{1}{5}\\
 \Leftrightarrow {x^2} + \frac{1}{3}x - \left( {{x^2} - \frac{5}{2}x + 8x - 20} \right) = \frac{1}{5}\\
 \Leftrightarrow  - \frac{{31}}{6}x + 20 = \frac{1}{5}\\
 \Leftrightarrow  - \frac{{31}}{6}x =  - \frac{{99}}{5}\\
 \Leftrightarrow x = \frac{{594}}{{155}}\\
Vậy\,x = \frac{{594}}{{155}}\\
18)\\
\left( {3x + 7} \right)\left( {2x - 1} \right) - \left( {x - 2} \right)\left( {6x + 3} \right) = 52\\
 \Leftrightarrow 6{x^2} + 11x - 7 - \left( {6{x^2} - 9x - 6} \right) = 52\\
 \Leftrightarrow 20x = 53\\
 \Leftrightarrow x = \frac{{53}}{{20}}\\
Vậy\,x = \frac{{53}}{{20}}\\
19)\\
{\left( {4x + 3} \right)^2} - {\left( {4x - 3} \right)^2} - 5x - 2 = 0\\
 \Leftrightarrow \left( {4x + 3 - 4x + 3} \right)\left( {4x + 3 + 4x - 3} \right) - 5x - 2 = 0\\
 \Leftrightarrow 6.8x - 5x - 2 = 0\\
 \Leftrightarrow 43x = 2\\
 \Leftrightarrow x = \frac{2}{{43}}\\
Vậy\,x = \frac{2}{{43}}
\end{array}$

$\begin{array}{l}
20)\\
3{\left( {x - 2} \right)^2} + 9\left( {x - 1} \right) - 3\left( {{x^2} + x - 3} \right) = 12\\
 \Leftrightarrow 3\left( {{x^2} - 4x + 4} \right) + 9x - 9\\
 - 3{x^2} - 3x + 9 = 12\\
 \Leftrightarrow  - 12x + 9x - 3x + 12 = 12\\
 \Leftrightarrow 12 = 12\left( {tm} \right)
\end{array}$

Vậy đa thức có nghiệm với mọi x