Đáp án:
$\begin{array}{l}
a)3{\left( {x + 3} \right)^2} + {\left( {3x - 1} \right)^2} - 12\left( {x - 3} \right)\left( {x + 3} \right) = 43\\
\Leftrightarrow 3\left( {{x^2} + 6x + 9} \right) + 9{x^2} - 6x + 1\\
- 12\left( {{x^2} - 9} \right) = 43\\
\Leftrightarrow 12{x^2} + 12x + 28 - 12{x^2} + 108 = 43\\
\Leftrightarrow 12x = - 93\\
\Leftrightarrow x = \dfrac{{ - 93}}{{12}}\\
Vậy\,x = \dfrac{{ - 93}}{{12}}\\
g)\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right) - x\left( {x + 5} \right)\left( {x - 5} \right) = 13\\
\Leftrightarrow {x^3} - 8 - x\left( {{x^2} - 25} \right) = 13\\
\Leftrightarrow {x^3} - 8 - {x^3} + 25x = 13\\
\Leftrightarrow 25x = 21\\
\Leftrightarrow x = \dfrac{{21}}{{25}}\\
Vậy\,x = \dfrac{{21}}{{25}}\\
h){\left( {x - 1} \right)^3} - \left( {x + 3} \right)\left( {{x^2} - 3x + 9} \right) + 3\left( {{x^2} - 4} \right) = 2\\
\Leftrightarrow {x^3} - 3{x^2} + 3x - 1 - {x^3} - 27 + 3{x^2} - 12 = 2\\
\Leftrightarrow 3x = 42\\
\Leftrightarrow x = 14\\
Vậy\,x = 14\\
i)\left( {x + 4} \right)\left( {{x^2} - 4x + 16} \right) - x\left( {x - 3} \right)\left( {x + 3} \right) = 45\\
\Leftrightarrow {x^3} + 64 - x\left( {{x^2} - 9} \right) = 45\\
\Leftrightarrow {x^3} + 64 - {x^3} + 9x = 45\\
\Leftrightarrow 9x = - 19\\
\Leftrightarrow x = - \dfrac{{19}}{9}\\
Vậy\,x = - \dfrac{{19}}{9}\\
j){\left( {x - 3} \right)^3} - \left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right) + 6\left( {x - 1} \right)\left( {x + 1} \right) = 17\\
\Leftrightarrow {x^3} - 9{x^2} + 27x - 27 - {x^3} + 8 + 6{x^2} - 6 = 17\\
\Leftrightarrow - 3{x^2} + 27x - 42 = 0\\
\Leftrightarrow {x^2} - 9x + 14 = 0\\
\Leftrightarrow \left( {x - 2} \right)\left( {x - 7} \right) = 0\\
\Leftrightarrow x = 2;x = 7\\
Vậy\,x = 2;x = 7\\
k)\\
{\left( {3x - 1} \right)^2} - \left( {2x - 5} \right)\left( {2x + 5} \right) - 5{x^2} = 15\\
\Leftrightarrow 9{x^2} - 6x + 1 - 4{x^2} + 25 - 5{x^2} = 15\\
\Leftrightarrow 6x = 11\\
\Leftrightarrow x = \dfrac{{11}}{6}\\
Vậy\,x = \dfrac{{11}}{6}
\end{array}$
