Đáp án:
e. \(\left[ \begin{array}{l}
x = - 1\\
x = - \dfrac{2}{3}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.x\left( {x - 6} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x - 6 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 0\\
x = 6
\end{array} \right.\\
c.\left( {6 - 3x} \right)\left( { - 2x + 5} \right) = 0\\
\to \left[ \begin{array}{l}
6 - 3x = 0\\
5 - 2x = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2\\
x = \dfrac{5}{2}
\end{array} \right.\\
d.x\left( {x + 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x + 1 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 0\\
x = - 1
\end{array} \right.\\
e.3{x^2} + 3x + 2x + 2 = 0\\
\to 3x\left( {x + 1} \right) + 2\left( {x + 1} \right) = 0\\
\to \left( {x + 1} \right)\left( {3x + 2} \right) = 0\\
\to \left[ \begin{array}{l}
x + 1 = 0\\
3x + 2 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 1\\
x = - \dfrac{2}{3}
\end{array} \right.
\end{array}\)
