Đáp án:
$\begin{array}{l}
a)8\left( {x - \dfrac{1}{2}} \right)\left( {{x^2} + \dfrac{1}{2}x + \dfrac{1}{4}} \right) - x\left( {1 + 8{x^2}} \right) + 2 = 0\\
\Leftrightarrow 8.\left( {{x^3} - \dfrac{1}{8}} \right) - x - 8{x^3} + 2 = 0\\
\Leftrightarrow 8{x^3} - 1 - x - 8{x^3} + 2 = 0\\
\Leftrightarrow x = 1\\
Vậy\,x = 1\\
c)\left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right) - x\left( {{x^2} - 2} \right) = 15\\
\Leftrightarrow {x^3} + 8 - {x^3} + 2x = 15\\
\Leftrightarrow 2x = 7\\
x = \dfrac{7}{2}\\
Vậy\,x = \dfrac{7}{2}\\
d){\left( {x - 3} \right)^3} - \left( {x - 3} \right)\left( {{x^2} + 3x + 9} \right)\\
+ 9{\left( {x + 1} \right)^2} = 15\\
\Leftrightarrow {x^3} - 9{x^2} + 27x - 27 - {x^3} + 27\\
+ 9{x^2} + 18x + 9 = 15\\
\Leftrightarrow 45x = 6\\
\Leftrightarrow x = \dfrac{2}{{15}}\\
Vậy\,x = \dfrac{2}{{15}}
\end{array}$
Sửa phần b:
$\begin{array}{l}
b){\left( {x - 1} \right)^3} + \left( {2 - x} \right)\left( {4 + 2x + {x^2}} \right) + 3x\left( {x + 2} \right) = 16\\
\Leftrightarrow {x^3} - 3{x^2} + 3x - 1 + {2^3} - {x^3} + 3{x^2} + 6x = 16\\
\Leftrightarrow 9x = 9\\
\Leftrightarrow x = 1\\
Vậy\,x = 1
\end{array}$
