Đáp án:
$\begin{array}{l}
{\left( {x + 3} \right)^2} - \left( {x + 4} \right)\left( {x - 4} \right) = 14\\
\Leftrightarrow {x^2} + 6x + 9 - \left( {{x^2} - 16} \right) = 14\\
\Leftrightarrow {x^2} + 6x + 9 - {x^2} + 16 = 14\\
\Leftrightarrow 5x = - 11\\
\Leftrightarrow x = - \dfrac{{11}}{5}\\
\text{Vậy}\,x = \dfrac{{ - 11}}{5}\\
a)3{x^2}\left( {{x^3} - 2xy + 4} \right) - 3\left( {{x^5} - 2{x^3}y + {x^2}} \right)\\
= 3{x^5} - 6{x^3}y + 12{x^2} - 3{x^5} + 6{x^3}y - 3{x^2}\\
= 9{x^2}\\
b){\left( {x + 3} \right)^2} - x\left( {x + 6} \right)\\
= {x^2} + 6x + 9 - {x^2} - 6x\\
= 9\\
c)\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right) - x\left( {{x^2} - 5} \right)\\
= {x^3} - {2^3} - {x^3} + 5x\\
= 5x - 8
\end{array}$
