Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
\left( {2x + 1} \right)\left( {4{x^2} - 2x + 1} \right) - 8x.\left( {{x^2} + 2} \right) = 17\\
 \Leftrightarrow \left( {2x + 1} \right).\left[ {{{\left( {2x} \right)}^2} - 2x.1 + {1^2}} \right] - \left( {8{x^3} + 16x} \right) = 17\\
 \Leftrightarrow \left[ {{{\left( {2x} \right)}^3} + {1^3}} \right] - \left( {8{x^3} + 16x} \right) = 17\\
 \Leftrightarrow 8{x^3} + 1 - 8{x^3} - 16x = 17\\
 \Leftrightarrow  - 16x = 16\\
 \Leftrightarrow x =  - 1\\
b,\\
{x^3} + 27 + \left( {x + 3} \right)\left( {x + 9} \right) = 0\\
 \Leftrightarrow \left( {{x^3} + {3^3}} \right) + \left( {x + 3} \right)\left( {x + 9} \right) = 0\\
 \Leftrightarrow \left( {x + 3} \right).\left( {{x^2} - x.3 + {3^2}} \right) + \left( {x + 3} \right)\left( {x + 9} \right) = 0\\
 \Leftrightarrow \left( {x + 3} \right).\left( {{x^2} - 3x + 9} \right) + \left( {x + 3} \right)\left( {x + 9} \right) = 0\\
 \Leftrightarrow \left( {x + 3} \right).\left[ {\left( {{x^2} - 3x + 9} \right) + \left( {x + 9} \right)} \right] = 0\\
 \Leftrightarrow \left( {x + 3} \right).\left( {{x^2} - 2x + 18} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x + 3 = 0\\
{x^2} - 2x + 18 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - 3\\
{\left( {x - 1} \right)^2} + 17 = 0\,\,\,\,\left( {vn} \right)
\end{array} \right.\\
 \Leftrightarrow x =  - 3\\
c,\\
4{x^2} - 25 - \left( {2x - 5} \right)\left( {2x + 7} \right) = 0\\
 \Leftrightarrow \left( {2x - 5} \right)\left( {2x + 5} \right) - \left( {2x - 5} \right)\left( {2x + 7} \right) = 0\\
 \Leftrightarrow \left( {2x - 5} \right).\left[ {\left( {2x + 5} \right) - \left( {2x + 7} \right)} \right] = 0\\
 \Leftrightarrow \left( {2x - 5} \right).\left( { - 2} \right) = 0\\
 \Leftrightarrow 2x - 5 = 0\\
 \Leftrightarrow x = \frac{5}{2}\\
d,\\
{x^3} + 27 + \left( {x + 3} \right)\left( {x - 9} \right) = 0\\
 \Leftrightarrow \left( {{x^3} + {3^3}} \right) + \left( {x + 3} \right)\left( {x - 9} \right) = 0\\
 \Leftrightarrow \left( {x + 3} \right).\left( {{x^2} - x.3 + {3^2}} \right) + \left( {x + 3} \right)\left( {x - 9} \right) = 0\\
 \Leftrightarrow \left( {x + 3} \right).\left( {{x^2} - 3x + 9} \right) + \left( {x + 3} \right)\left( {x - 9} \right) = 0\\
 \Leftrightarrow \left( {x + 3} \right).\left[ {\left( {{x^2} - 3x + 9} \right) + \left( {x - 9} \right)} \right] = 0\\
 \Leftrightarrow \left( {x + 3} \right).\left( {{x^2} - 2x} \right) = 0\\
 \Leftrightarrow \left( {x + 3} \right).x\left( {x - 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x + 3 = 0\\
x = 0\\
x - 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x =  - 3\\
x = 0\\
x = 2
\end{array} \right.\\
e,\\
8{x^3} - 12{x^2} + 6x - 1 = 0\\
 \Leftrightarrow {\left( {2x} \right)^3} - 3.{\left( {2x} \right)^2}.1 + 3.2x.1 - {1^3} = 0\\
 \Leftrightarrow {\left( {2x - 1} \right)^3} = 0\\
 \Leftrightarrow 2x - 1 = 0\\
 \Leftrightarrow x = \frac{1}{2}
\end{array}\)