Đáp án:
$\begin{array}{l}
d)\left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right) - x\left( {x - 3} \right)\left( {x + 3} \right) = 26\\
\Rightarrow {x^3} + {2^3} - x\left( {{x^2} - {3^2}} \right) = 26\\
\Rightarrow {x^3} + 8 - {x^3} + 9x - 26 = 0\\
\Rightarrow 9x - 18 = 0\\
\Rightarrow x = 2\\
Vậy\,x = 2\\
e)\left( {3x + 2} \right)\left( {3x - 2} \right) - {\left( {3x - 4} \right)^2} = 28\\
\Rightarrow {\left( {3x} \right)^2} - {2^2} - \left( {9{x^2} - 24x + 16} \right) = 28\\
\Rightarrow 9{x^2} - 4 - 9{x^2} + 24x - 16 - 28 = 0\\
\Rightarrow 24x - 48 = 0\\
\Rightarrow x = 2\\
Vậy\,x = 2\\
g)x\left( {x - 2} \right)\left( {x + 2} \right) - \left( {x - 3} \right)\left( {{x^2} + 3x + 9} \right) + 1 = 0\\
\Rightarrow x\left( {{x^2} - 4} \right) - \left( {{x^3} - {3^3}} \right) + 1 = 0\\
\Rightarrow {x^3} - 4x - {x^3} + 27 + 1 = 0\\
\Rightarrow 4x = 28\\
\Rightarrow x = 7\\
Vậy\,x = 7\\
h)\frac{{2\left( {5x + 2} \right)}}{9} - 1 = \frac{{4\left( {33 + 2x} \right)}}{5} - \frac{{5\left( {1 - 11x} \right)}}{9}\\
\Rightarrow \frac{{10x + 4}}{9} + \frac{{5 - 55x}}{9} = 1 + \frac{{132 + 8x}}{5}\\
\Rightarrow \frac{{9 - 45x}}{9} = \frac{{137 + 8x}}{5}\\
\Rightarrow 9.\left( {137 + 8x} \right) = 5.\left( {9 - 45x} \right)\\
\Rightarrow 1233 + 72x = 45 - 225x\\
\Rightarrow 72x + 225x = 45 - 1233\\
\Rightarrow 297x = - 1188\\
\Rightarrow x = - 4\\
Vậy\,x = - 4\\
i)\frac{{2\left( {x - 4} \right)}}{3} + \frac{{3x + 13}}{8} = \frac{{2\left( {2x - 3} \right)}}{5} - 7\\
\Rightarrow \frac{{40.\left( {2x - 8} \right) + 15.\left( {3x + 13} \right)}}{{120}} = \frac{{24.\left( {4x - 6} \right) - 7.120}}{{120}}\\
\Rightarrow 80x - 320 + 45x + 195 = 96x - 144 - 840\\
\Rightarrow 80x + 45x - 96x = 320 - 195 - 144 - 840\\
\Rightarrow 29x = - 859\\
\Rightarrow x = \frac{{ - 859}}{{29}}\\
Vậy\,x = \frac{{ - 859}}{{29}}
\end{array}$
