Đáp án:

\(\begin{array}{l}
a,\,\,\left[ \begin{array}{l}
x = 0\\
x = \sqrt {13} \\
x =  - \sqrt {13} 
\end{array} \right.\\
b,\,\,\,\left[ \begin{array}{l}
x = 2000\\
x = \dfrac{1}{5}
\end{array} \right.\\
c,\,\,\left[ \begin{array}{l}
x = 2\\
x =  - \dfrac{3}{2}
\end{array} \right.\\
d,\,\,\left[ \begin{array}{l}
x = 0\\
x =  - \dfrac{1}{5}
\end{array} \right.\\
d,\,\,\left[ \begin{array}{l}
x =  - 1\\
x = 0
\end{array} \right.\\
e,\,\,x = 0\\
f,\,\,\,\left[ \begin{array}{l}
x = 2\\
x =  - 1
\end{array} \right.\\
g,\,\,\,\left[ \begin{array}{l}
x = 3\\
x = \dfrac{1}{5}
\end{array} \right.\\
h,\,\,\,\left[ \begin{array}{l}
x = 3\\
x = 2\\
x =  - 2
\end{array} \right.
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
a,\,\,{x^3} - 13x = 0\\
 \Leftrightarrow x.\left( {{x^2} - 13} \right) = 0\\
 \Leftrightarrow x\left( {{x^2} - {{\sqrt {13} }^2}} \right) = 0\\
 \Leftrightarrow x.\left( {x - \sqrt {13} } \right).\left( {x + \sqrt {13} } \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x - \sqrt {13}  = 0\\
x + \sqrt {13}  = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = \sqrt {13} \\
x =  - \sqrt {13} 
\end{array} \right.\\
b,\,\,\,5x\left( {x - 2000} \right) - x + 2000 = 0\\
 \Leftrightarrow 5x\left( {x - 2000} \right) - \left( {x - 2000} \right) = 0\\
 \Leftrightarrow \left( {x - 2000} \right)\left( {5x - 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 2000 = 0\\
5x - 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 2000\\
x = \dfrac{1}{5}
\end{array} \right.\\
c,\,\,2x\left( {x - 2} \right) + 3.\left( {x - 2} \right) = 0\\
 \Leftrightarrow \left( {x - 2} \right)\left( {2x + 3} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 2 = 0\\
2x + 3 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x =  - \dfrac{3}{2}
\end{array} \right.\\
d,\,\,x + 5{x^2} = 0\\
 \Leftrightarrow x.\left( {1 + 5x} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
1 + 5x = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x =  - \dfrac{1}{5}
\end{array} \right.\\
d,\,\,x + 1 = {\left( {x + 1} \right)^2}\\
 \Leftrightarrow {\left( {x + 1} \right)^2} - \left( {x + 1} \right) = 0\\
 \Leftrightarrow \left( {x + 1} \right).\left[ {\left( {x + 1} \right) - 1} \right] = 0\\
 \Leftrightarrow \left( {x + 1} \right).x = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x + 1 = 0\\
x = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x =  - 1\\
x = 0
\end{array} \right.\\
e,\,\,{x^3} + x = 0\\
 \Leftrightarrow x.\left( {{x^2} + 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
{x^2} + 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
{x^2} =  - 1\,\,\,\left( L \right)
\end{array} \right. \Leftrightarrow x = 0\\
f,\,\,\,x\left( {x - 2} \right) + x - 2 = 0\\
 \Leftrightarrow x\left( {x - 2} \right) + \left( {x - 2} \right) = 0\\
 \Leftrightarrow \left( {x - 2} \right)\left( {x + 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 2 = 0\\
x + 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 2\\
x =  - 1
\end{array} \right.\\
g,\,\,\,5x\left( {x - 3} \right) - x + 3 = 0\\
 \Leftrightarrow 5x\left( {x - 3} \right) - \left( {x - 3} \right) = 0\\
 \Leftrightarrow \left( {x - 3} \right)\left( {5x - 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 3 = 0\\
5x - 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = \dfrac{1}{5}
\end{array} \right.\\
h,\,\,\,{x^2}\left( {x - 3} \right) + 12 - 4x = 0\\
 \Leftrightarrow {x^2}\left( {x - 3} \right) + \left( { - 4x + 12} \right) = 0\\
 \Leftrightarrow {x^2}\left( {x - 3} \right) - 4.\left( {x - 3} \right) = 0\\
 \Leftrightarrow \left( {x - 3} \right)\left( {{x^2} - 4} \right) = 0\\
 \Leftrightarrow \left( {x - 3} \right).\left( {{x^2} - {2^2}} \right) = 0\\
 \Leftrightarrow \left( {x - 3} \right)\left( {x - 2} \right)\left( {x + 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 3 = 0\\
x - 2 = 0\\
x + 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = 2\\
x =  - 2
\end{array} \right.
\end{array}\)