Đáp án:
$\begin{array}{l}
a){\left( {x - 1} \right)^3} + 5\left( {2 - x} \right) + 3x\left( {x + 2} \right) = 17\\
\Leftrightarrow {x^3} - 3{x^2} + 3x - 1 + 10 - 5x\\
+ 3{x^2} + 6x = 17\\
\Leftrightarrow {x^3} + 4x - 8 = 0\\
\Leftrightarrow x = 1,36\\
Vậy\,x = 1,36\\
b){\left( {x + 2} \right)^3} - x\left( {{x^2} + 6x} \right) = 15\\
\Leftrightarrow {x^3} + 6{x^2} + 12x + 8 - {x^3} - 6{x^2} = 15\\
\Leftrightarrow 12x = 15 - 8\\
\Leftrightarrow 12x = 7\\
\Leftrightarrow x = \dfrac{7}{{12}}\\
Vậy\,x = \dfrac{7}{{12}}\\
c){\left( {x - 3} \right)^3} - \left( {{x^3} - 7} \right) + 9{\left( {x + 1} \right)^2} = 15\\
\Leftrightarrow {x^3} - 9{x^2} + 27x - 27\\
- {x^3} + 7 + 9{x^2} + 18x + 9 = 15\\
\Leftrightarrow 45x = 26\\
\Leftrightarrow x = \dfrac{{26}}{{45}}\\
Vậy\,x = \dfrac{{26}}{{45}}\\
d)x\left( {x - 5} \right)\left( {x + 5} \right) - {\left( {x - 1} \right)^3} + 2 = 3\\
\Leftrightarrow x\left( {{x^2} - 25} \right) - {x^3} + 3{x^2} - 3x + 1 + 2 - 3 = 0\\
\Leftrightarrow {x^3} - 25x - {x^3} + 3{x^2} - 3x = 0\\
\Leftrightarrow 3{x^2} - 28x = 0\\
\Leftrightarrow x\left( {3x - 28} \right) = 0\\
\Leftrightarrow x = 0;x = \dfrac{{28}}{3}\\
Vậy\,x = 0;x = \dfrac{{28}}{3}
\end{array}$
