Đáp án:

\(\begin{array}{l}
a,\\
\left[ \begin{array}{l}
x = 5\\
x = 1\\
x =  - 1
\end{array} \right.\\
b,\\
\left[ \begin{array}{l}
x = 3\\
x = 0
\end{array} \right.\\
c,\\
\left[ \begin{array}{l}
x =  - 8\\
x = 0\\
x =  - 1
\end{array} \right.\\
d,\\
x =  - 3
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
a,\\
{x^2}\left( {x - 5} \right) + 5 - x = 0\\
 \Leftrightarrow {x^2}\left( {x - 5} \right) - \left( {x - 5} \right) = 0\\
 \Leftrightarrow \left( {x - 5} \right)\left( {{x^2} - 1} \right) = 0\\
 \Leftrightarrow \left( {x - 5} \right)\left( {{x^2} - {1^2}} \right) = 0\\
 \Leftrightarrow \left( {x - 5} \right)\left( {x - 1} \right)\left( {x + 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 5 = 0\\
x - 1 = 0\\
x + 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 5\\
x = 1\\
x =  - 1
\end{array} \right.\\
b,\\
3{x^4} - 9{x^3} =  - 9{x^2} + 27x\\
 \Leftrightarrow 3{x^3}.\left( {x - 3} \right) =  - 9x.\left( {x - 3} \right)\\
 \Leftrightarrow 3{x^3}\left( {x - 3} \right) + 9x\left( {x - 3} \right) = 0\\
 \Leftrightarrow \left( {x - 3} \right).\left( {3{x^3} + 9x} \right) = 0\\
 \Leftrightarrow \left( {x - 3} \right).3x.\left( {{x^2} + 3} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 3 = 0\\
x = 0\\
{x^2} + 3 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = 0\\
{x^2} =  - 3\,\,\,\,\left( {L,\,\,{x^2} \ge 0} \right)
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = 0
\end{array} \right.\\
c,\\
{x^2}\left( {x + 8} \right) + {x^2} =  - 8x\\
 \Leftrightarrow {x^2}\left( {x + 8} \right) + {x^2} + 8x = 0\\
 \Leftrightarrow {x^2}\left( {x + 8} \right) + x.\left( {x + 8} \right) = 0\\
 \Leftrightarrow \left( {x + 8} \right)\left( {{x^2} + x} \right) = 0\\
 \Leftrightarrow \left( {x + 8} \right).x.\left( {x + 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x + 8 = 0\\
x = 0\\
x + 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x =  - 8\\
x = 0\\
x =  - 1
\end{array} \right.\\
d,\\
\left( {x + 3} \right)\left( {{x^2} - 3x + 5} \right) = {x^2} + 3x\\
 \Leftrightarrow \left( {x + 3} \right)\left( {{x^2} - 3x + 5} \right) = x\left( {x + 3} \right)\\
 \Leftrightarrow \left( {x + 3} \right)\left( {{x^2} - 3x + 5} \right) - x\left( {x + 3} \right) = 0\\
 \Leftrightarrow \left( {x + 3} \right).\left[ {\left( {{x^2} - 3x + 5} \right) - x} \right] = 0\\
 \Leftrightarrow \left( {x + 3} \right)\left( {{x^2} - 4x + 5} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x + 3 = 0\\
{x^2} - 4x + 5 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x =  - 3\\
{x^2} - 4x + 4 =  - 1
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - 3\\
{x^2} - 2.x.2 + {2^2} =  - 1
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - 3\\
{\left( {x - 2} \right)^2} =  - 1\,\,\,\,\left( {L,\,\,\,{{\left( {x - 2} \right)}^2} \ge 0} \right)
\end{array} \right.\\
 \Leftrightarrow x =  - 3
\end{array}\)