Đáp án:

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

Giải thích các bước giải:

\(\begin{array}{l}
1)x\left( {x - 3} \right) = 0\\
 \to \left[ \begin{array}{l}
x = 0\\
x - 3 = 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 0\\
x = 3
\end{array} \right.\\
4)2x\left( {x - 3} \right) - 4\left( {x - 3} \right) = 0\\
 \to \left( {x - 3} \right)\left( {2x - 4} \right) = 0\\
 \to \left[ \begin{array}{l}
x - 3 = 0\\
2x - 4 = 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 3\\
x = 2
\end{array} \right.\\
7)8x\left( {x - 5} \right) - 2\left( {x - 5} \right) = 0\\
 \to \left( {8x - 2} \right)\left( {x - 5} \right) = 0\\
 \to \left[ \begin{array}{l}
8x - 2 = 0\\
x - 5 = 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = \dfrac{1}{4}\\
x = 5
\end{array} \right.\\
2)2x - 8{x^3} = 0\\
 \to 2x\left( {1 - 4{x^2}} \right) = 0\\
 \to \left[ \begin{array}{l}
x = 0\\
\left( {1 - 2x} \right)\left( {1 + 2x} \right) = 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 0\\
x = \dfrac{1}{2}\\
x =  - \dfrac{1}{2}
\end{array} \right.\\
5)\left( {x - 2} \right) - 16{\left( {x - 2} \right)^3} = 0\\
 \to \left( {x - 2} \right)\left( {1 - 16{{\left( {x - 2} \right)}^2}} \right) = 0\\
 \to \left[ \begin{array}{l}
x - 2 = 0\\
\left( {1 - 4\left( {x - 2} \right)} \right)\left( {1 + 4\left( {x - 2} \right)} \right) = 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 2\\
4x - 8 = 1\\
4x - 8 =  - 1
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 2\\
x = \dfrac{9}{4}\\
x = \dfrac{7}{4}
\end{array} \right.\\
8)\left( {{x^2} + 1} \right)\left( {x - 2} \right) + 2\left( {x - 2} \right) = 0\\
 \to \left( {x - 2} \right)\left( {{x^2} + 1 + 2} \right) = 0\\
 \to x - 2 = 0\left( {do:{x^2} + 3 > 0\forall x} \right)\\
 \to x = 2\\
3)5{x^2} = 15{x^3}\\
 \to 15{x^3} - 5{x^2} = 0\\
 \to 5{x^2}\left( {3x - 1} \right) = 0\\
 \to \left[ \begin{array}{l}
x = 0\\
x = \dfrac{1}{3}
\end{array} \right.\\
6)x\left( {x + 3} \right) - 5\left( {x - 3} \right)\left( {x + 3} \right) = 0\\
 \to \left( {x + 3} \right)\left( {x - 5\left( {x - 3} \right)} \right) = 0\\
 \to \left[ \begin{array}{l}
x + 3 = 0\\
x - 5x + 15 = 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x =  - 3\\
x = \dfrac{{15}}{4}
\end{array} \right.\\
9)\left( {5 - x} \right)\left( {x - 2} \right) + {x^2} - 4 = 0\\
 \to \left( {5 - x} \right)\left( {x - 2} \right) + \left( {x - 2} \right)\left( {x + 2} \right) = 0\\
 \to \left( {x - 2} \right)\left( {5 - x + x + 2} \right) = 0\\
 \to x - 2 = 0\\
 \to x = 2
\end{array}\)