Đáp án:

\(\begin{array}{l}
g,\\
\left[ \begin{array}{l}
x = 4\\
x = 1\\
x =  - 1
\end{array} \right.\\
h,\\
\left[ \begin{array}{l}
x = 0\\
x = 1
\end{array} \right.\\
k,\\
\left[ \begin{array}{l}
x = 0\\
x = \dfrac{1}{2}\\
x =  - \dfrac{1}{2}
\end{array} \right.\\
l,\\
\left[ \begin{array}{l}
x = 1\\
x = \dfrac{2}{5}
\end{array} \right.\\
m,\\
\left[ \begin{array}{l}
x = 3\\
x = \dfrac{2}{5}
\end{array} \right.\\
n,\\
\left[ \begin{array}{l}
x =  - 1\\
x = 2\\
x =  - 2
\end{array} \right.
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
g,\\
{x^2}\left( {x - 4} \right) - \left( {x - 4} \right) = 0\\
 \Leftrightarrow \left( {x - 4} \right)\left( {{x^2} - 1} \right) = 0\\
 \Leftrightarrow \left( {x - 4} \right)\left( {{x^2} - {1^2}} \right) = 0\\
 \Leftrightarrow \left( {x - 4} \right)\left( {x - 1} \right)\left( {x + 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 4 = 0\\
x - 1 = 0\\
x + 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 4\\
x = 1\\
x =  - 1
\end{array} \right.\\
h,\\
{x^3} - 2{x^2} =  - x\\
 \Leftrightarrow {x^3} - 2{x^2} + x = 0\\
 \Leftrightarrow x.\left( {{x^2} - 2x + 1} \right) = 0\\
 \Leftrightarrow x.\left( {{x^2} - 2.x.1 + {1^2}} \right) = 0\\
 \Leftrightarrow x.{\left( {x - 1} \right)^2} = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
{\left( {x - 1} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 1
\end{array} \right.\\
k,\\
{x^3} = \dfrac{1}{4}x\\
 \Leftrightarrow {x^3} - \dfrac{1}{4}x = 0\\
 \Leftrightarrow x.\left( {{x^2} - \dfrac{1}{4}} \right) = 0\\
 \Leftrightarrow x.\left[ {{x^2} - {{\left( {\dfrac{1}{2}} \right)}^2}} \right] = 0\\
 \Leftrightarrow x.\left( {x - \dfrac{1}{2}} \right).\left( {x + \dfrac{1}{2}} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x - \dfrac{1}{2} = 0\\
x + \dfrac{1}{2} = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = \dfrac{1}{2}\\
x =  - \dfrac{1}{2}
\end{array} \right.\\
l,\\
5x\left( {x - 1} \right) - 2x + 2 = 0\\
 \Leftrightarrow 5x\left( {x - 1} \right) - \left( {2x - 2} \right) = 0\\
 \Leftrightarrow 5x\left( {x - 1} \right) - 2\left( {x - 1} \right) = 0\\
 \Leftrightarrow \left( {x - 1} \right)\left( {5x - 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
5x - 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = \dfrac{2}{5}
\end{array} \right.\\
m,\\
\left( {5x - 1} \right)\left( {x - 3} \right) + 3 - x = 0\\
 \Leftrightarrow \left( {5x - 1} \right)\left( {x - 3} \right) - \left( {x - 3} \right) = 0\\
 \Leftrightarrow \left( {x - 3} \right).\left[ {\left( {5x - 1} \right) - 1} \right] = 0\\
 \Leftrightarrow \left( {x - 3} \right)\left( {5x - 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 3 = 0\\
5x - 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = \dfrac{2}{5}
\end{array} \right.\\
n,\\
{x^3} + {x^2} - 4x - 4 = 0\\
 \Leftrightarrow \left( {{x^3} + {x^2}} \right) + \left( { - 4x - 4} \right) = 0\\
 \Leftrightarrow {x^2}\left( {x + 1} \right) - 4\left( {x + 1} \right) = 0\\
 \Leftrightarrow \left( {x + 1} \right)\left( {{x^2} - 4} \right) = 0\\
 \Leftrightarrow \left( {x + 1} \right)\left( {{x^2} - {2^2}} \right) = 0\\
 \Leftrightarrow \left( {x + 1} \right)\left( {x - 2} \right)\left( {x + 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x + 1 = 0\\
x - 2 = 0\\
x + 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x =  - 1\\
x = 2\\
x =  - 2
\end{array} \right.
\end{array}\)