Đáp án:

\(\begin{array}{l}
1,\,\,\,\,\left[ \begin{array}{l}
x = 0\\
x = \dfrac{1}{2}
\end{array} \right.\\
2,\,\,\,\,\left[ \begin{array}{l}
x = 0\\
x = 9
\end{array} \right.\\
3,\,\,\,\left[ \begin{array}{l}
x = 0\\
x = \dfrac{{\sqrt {30} }}{{15}}\\
x =  - \dfrac{{\sqrt {30} }}{{15}}
\end{array} \right.\\
4,\,\,\,\,\left[ \begin{array}{l}
x =  - 4\\
x = \dfrac{5}{2}
\end{array} \right.\\
5,\,\,\,\,\left[ \begin{array}{l}
x =  - \dfrac{4}{3}\\
x = 0
\end{array} \right.\\
6,\,\,\,\left[ \begin{array}{l}
x =  - 3\\
x = 0\\
x = \dfrac{2}{9}
\end{array} \right.\\
7,\,\,\,\left[ \begin{array}{l}
x = 4\\
x = \dfrac{5}{2}
\end{array} \right.\\
8,\,\,\,\,\,\left[ \begin{array}{l}
x = \dfrac{4}{3}\\
x = 0
\end{array} \right.\\
9,\,\,\,\left[ \begin{array}{l}
x =  - 3\\
x = 0\\
x = \dfrac{2}{9}
\end{array} \right.\\
10,\,\,\,\left[ \begin{array}{l}
x = 4\\
x = 3\\
x =  - 3
\end{array} \right.
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\,\,\,\,4{x^3} - 2{x^2} = 0\\
 \Leftrightarrow 2{x^2}.\left( {2x - 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
{x^2} = 0\\
2x - 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = \dfrac{1}{2}
\end{array} \right.\\
2,\,\,\,\,{x^3} - 9{x^2} = 0\\
 \Leftrightarrow {x^2}.\left( {x - 9} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
{x^2} = 0\\
x - 9 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 9
\end{array} \right.\\
3,\,\,\,15{x^4} - 2{x^2} = 0\\
 \Leftrightarrow {x^2}.\left( {15{x^2} - 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
{x^2} = 0\\
15{x^2} - 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
{x^2} = \dfrac{2}{{15}}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = \sqrt {\dfrac{2}{{15}}} \\
x =  - \sqrt {\dfrac{2}{{15}}} 
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = \dfrac{{\sqrt {30} }}{{15}}\\
x =  - \dfrac{{\sqrt {30} }}{{15}}
\end{array} \right.\\
4,\,\,\,\,5\left( {x + 4} \right) - 2x\left( {x + 4} \right) = 0\\
 \Leftrightarrow \left( {x + 4} \right).\left( {5 - 2x} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x + 4 = 0\\
5 - 2x = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x =  - 4\\
x = \dfrac{5}{2}
\end{array} \right.\\
5,\,\,\,\,5x\left( {3x + 4} \right) - 2x\left( {3x + 5} \right) = 0\\
 \Leftrightarrow \left( {3x + 4} \right).\left( {5x - 2x} \right) = 0\\
 \Leftrightarrow \left( {3x + 4} \right).3x = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
3x + 4 = 0\\
3x = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x =  - \dfrac{4}{3}\\
x = 0
\end{array} \right.\\
6,\,\,\,9{x^2}\left( {x + 3} \right) - 2x\left( {x + 3} \right) = 0\\
 \Leftrightarrow \left( {x + 3} \right).\left( {9{x^2} - 2x} \right) = 0\\
 \Leftrightarrow \left( {x + 3} \right).x.\left( {9x - 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x + 3 = 0\\
x = 0\\
9x - 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x =  - 3\\
x = 0\\
x = \dfrac{2}{9}
\end{array} \right.\\
7,\,\,\,5\left( {x - 4} \right) - 2x\left( {x - 4} \right) = 0\\
 \Leftrightarrow \left( {x - 4} \right)\left( {5 - 2x} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 4 = 0\\
5 - 2x = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 4\\
x = \dfrac{5}{2}
\end{array} \right.\\
8,\,\,\,\,\,5x\left( {3x - 4} \right) - 2x.\left( {3x - 4} \right) = 0\\
 \Leftrightarrow \left( {3x - 4} \right).\left( {5x - 2x} \right) = 0\\
 \Leftrightarrow \left( {3x - 4} \right).3x = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
3x - 4 = 0\\
3x = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{4}{3}\\
x = 0
\end{array} \right.\\
9,\,\,\,9{x^2}\left( {x + 3} \right) - 2x\left( {x + 3} \right) = 0\\
 \Leftrightarrow \left( {x + 3} \right).\left( {9{x^2} - 2x} \right) = 0\\
 \Leftrightarrow \left( {x + 3} \right).x.\left( {9x - 2} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x + 3 = 0\\
x = 0\\
9x - 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x =  - 3\\
x = 0\\
x = \dfrac{2}{9}
\end{array} \right.\\
10,\,\,\,9\left( {x - 4} \right) - {x^2}\left( {x - 4} \right) = 0\\
 \Leftrightarrow \left( {x - 4} \right).\left( {9 - {x^2}} \right) = 0\\
 \Leftrightarrow \left( {x - 4} \right).\left( {{3^2} - {x^2}} \right) = 0\\
 \Leftrightarrow \left( {x - 4} \right)\left( {3 - x} \right)\left( {3 + x} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 4 = 0\\
3 - x = 0\\
3 + x = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 4\\
x = 3\\
x =  - 3
\end{array} \right.
\end{array}\)