Đáp án:
$\begin{array}{l}
a){\left( {x - 1} \right)^3} + \left( {2 - x} \right)\left( {4 + 2x + {x^2}} \right) + 3x\left( {x + 2} \right) = 17\\
\Leftrightarrow {x^3} - 3{x^2} + 3x - 1 + {2^3} - {x^3} + 3{x^2} + 6x = 17\\
\Leftrightarrow 9x = 10\\
\Leftrightarrow x = \dfrac{{10}}{9}\\
Vậy\,x = \dfrac{{10}}{9}\\
b)\left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right) - x\left( {{x^2} - 2} \right) = 15\\
\Leftrightarrow {x^3} + 8 - {x^3} + 2x = 15\\
\Leftrightarrow 2x = 7\\
\Leftrightarrow x = \dfrac{7}{2}\\
Vậy\,x = \dfrac{7}{2}\\
c){\left( {x - 3} \right)^3} - \left( {x - 3} \right)\left( {{x^2} + 3x + 9} \right) + 9{\left( {x + 1} \right)^2} = 15\\
\Leftrightarrow {x^3} - 9{x^2} + 27x - 27 - \left( {{x^3} - 27} \right)\\
+ 9\left( {{x^2} + 2x + 1} \right) = 15\\
\Leftrightarrow {x^3} - 9{x^2} + 27x - 27 - {x^3} + 27\\
+ 9{x^2} + 18x + 9 = 15\\
\Leftrightarrow 45x = 6\\
\Leftrightarrow x = \dfrac{2}{{15}}\\
Vậy\,x = \dfrac{2}{{15}}\\
d)x\left( {x - 5} \right)\left( {x + 5} \right) - \left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right) = 3\\
\Leftrightarrow x\left( {{x^2} - 25} \right) - \left( {{x^3} + 8} \right) = 3\\
\Leftrightarrow {x^3} - 25x - {x^3} - 8 = 3\\
\Leftrightarrow 25x = - 11\\
\Leftrightarrow x = - \dfrac{{11}}{{25}}\\
Vậy\,x = \dfrac{{ - 11}}{{25}}
\end{array}$
