Giải thích các bước giải:
\(\begin{array}{l}
1,\\
3x - 15 = 2x\left( {x - 5} \right)\\
\Leftrightarrow 3\left( {x - 5} \right) = 2x\left( {x - 5} \right)\\
\Leftrightarrow \left( {x - 5} \right)\left( {3 - 2x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 5 = 0\\
3 - 2x = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 5\\
x = \frac{3}{2}
\end{array} \right.\\
2,\\
\left( {x + 3} \right)\left( {{x^2} - 4} \right) = 0\\
\Leftrightarrow \left( {x + 3} \right)\left( {x - 2} \right)\left( {x + 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x + 3 = 0\\
x - 2 = 0\\
x + 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = - 3\\
x = \pm 2
\end{array} \right.\\
3,\\
\left( {x - 5} \right)\left( {{x^3} + 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 5 = 0\\
{x^3} + 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 5\\
x = - \sqrt[3]{2}
\end{array} \right.\\
4,\\
\left( {x + 1} \right)\left( {x + 4} \right) = \left( {2 - x} \right)\left( {x + 2} \right)\\
\Leftrightarrow {x^2} + 5x + 4 = 4 - {x^2}\\
\Leftrightarrow 2{x^2} + 5x = 0\\
\Leftrightarrow x\left( {2x + 5} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
2x + 5 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = - \frac{5}{2}
\end{array} \right.\\
5,\\
\left( {x + 2} \right)\left( {3 - 4x} \right) = {x^2} + 4x + 4\\
\Leftrightarrow \left( {x + 2} \right)\left( {3 - 4x} \right) = {\left( {x + 2} \right)^2}\\
\Leftrightarrow \left( {x + 2} \right).\left[ {\left( {3 - 4x} \right) - \left( {x + 2} \right)} \right] = 0\\
\Leftrightarrow \left( {x + 2} \right)\left( {1 - 5x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x + 2 = 0\\
1 - 5x = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = - 2\\
x = \frac{1}{5}
\end{array} \right.
\end{array}\)
