Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
{x^3} - {x^2} - 8x - 4 = 0\\
 \Leftrightarrow \left( {{x^3} + 2{x^2}} \right) - \left( {3{x^2} + 6x} \right) - \left( {2x + 4} \right) = 0\\
 \Leftrightarrow {x^2}\left( {x + 2} \right) - 3x\left( {x + 2} \right) - 2\left( {x + 2} \right) = 0\\
 \Leftrightarrow \left( {x + 2} \right).\left( {{x^2} - 3x - 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x + 2 = 0\\
{x^2} - 3x - 2 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - 2\\
{x^2} - 3x + \dfrac{9}{4} = \dfrac{{17}}{2}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - 2\\
{\left( {x - \dfrac{3}{2}} \right)^2} = \dfrac{{17}}{2}
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - 2\\
x = \dfrac{{3 \pm \sqrt {17} }}{2}
\end{array} \right.\\
b,\\
{\left( {{x^2} - 4} \right)^2} = 8x + 1\\
 \Leftrightarrow {\left( {{x^2} - 4} \right)^2} - {3^2} = 8x + 1 - {3^2}\\
 \Leftrightarrow \left( {{x^2} - 4 + 3} \right).\left( {{x^2} - 4 - 3} \right) = 8x - 8\\
 \Leftrightarrow \left( {{x^2} - 1} \right)\left( {{x^2} - 7} \right) = 8\left( {x - 1} \right)\\
 \Leftrightarrow \left( {x - 1} \right)\left( {x + 1} \right).\left( {{x^2} - 7} \right) = 8.\left( {x - 1} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
\left( {x + 1} \right).\left( {{x^2} - 7} \right) = 8
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
{x^3} + {x^2} - 7x - 15 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
\left( {{x^3} - 3{x^2}} \right) + \left( {4{x^2} - 12x} \right) + \left( {5x - 15} \right) = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
{x^2}\left( {x - 3} \right) + 4x\left( {x - 3} \right) + 5\left( {x - 3} \right) = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
\left( {x - 3} \right)\left( {{x^2} + 4x + 5} \right) = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x - 3 = 0\\
{x^2} + 4x + 5 = 0\,\,\,\,\,\,\,\,\,\,\,\,\left( {vn} \right)
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = 3
\end{array} \right.
\end{array}\) 

Bài 2:

ĐKXĐ: \(\left\{ \begin{array}{l}
x \ne 0\\
x \ne  \pm 3
\end{array} \right.\)

Ta có:

\(\begin{array}{l}
a,\\
B = \left( {\dfrac{1}{3} + \dfrac{3}{{{x^2} - 3x}}} \right):\left( {\dfrac{{{x^2}}}{{27 - 3{x^2}}} + \dfrac{1}{{x + 3}}} \right)\\
 = \dfrac{{\left( {{x^2} - 3x} \right) + 9}}{{3\left( {{x^2} - 3x} \right)}}:\left( {\dfrac{{{x^2}}}{{3.\left( {9 - {x^2}} \right)}} + \dfrac{1}{{x + 3}}} \right)\\
 = \dfrac{{{x^2} - 3x + 9}}{{3\left( {{x^2} - 3x} \right)}}:\left( {\dfrac{{{x^2}}}{{3.\left( {3 - x} \right)\left( {3 + x} \right)}} + \dfrac{1}{{x + 3}}} \right)\\
 = \dfrac{{{x^2} - 3x + 9}}{{3x\left( {x - 3} \right)}}:\dfrac{{{x^2} + 3.\left( {3 - x} \right)}}{{3\left( {3 - x} \right)\left( {3 + x} \right)}}\\
 = \dfrac{{{x^2} - 3x + 9}}{{3x\left( {x - 3} \right)}}:\dfrac{{{x^2} - 3x + 9}}{{3\left( {3 - x} \right)\left( {3 + x} \right)}}\\
 = \dfrac{{{x^2} - 3x + 9}}{{3x\left( {x - 3} \right)}}.\dfrac{{3.\left( {3 - x} \right)\left( {3 + x} \right)}}{{{x^2} - 3x + 9}}\\
 =  - \dfrac{{x + 3}}{x}\\
b,\\
B <  - 1 \Leftrightarrow  - \dfrac{{x + 3}}{x} <  - 1\\
 \Leftrightarrow \dfrac{{x + 3}}{x} > 1\\
 \Leftrightarrow \dfrac{{x + 3}}{x} - 1 > 0\\
 \Leftrightarrow \dfrac{{x + 3 - x}}{x} > 0\\
 \Leftrightarrow \dfrac{3}{x} > 0\\
 \Leftrightarrow x > 0
\end{array}\)

Kết hợp ĐKXĐ ta được \(\left\{ \begin{array}{l}
x > 0\\
x \ne 3
\end{array} \right.\) thì \(B <  - 1\)