Đáp án:
f) \(\left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x = \dfrac{4}{3}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)2x\left( {3x - 1} \right) - \left( {3x - 1} \right) = 0\\
\to \left( {3x - 1} \right)\left( {2x - 1} \right) = 0\\
\to \left[ \begin{array}{l}
3x - 1 = 0\\
2x - 1 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{1}{3}\\
x = \dfrac{1}{2}
\end{array} \right.\\
c)\left( {x - 1} \right)\left( {2x + 3} \right) + 2\left( {x - 1} \right) = 0\\
\to \left( {x - 1} \right)\left( {2x + 3 + 2} \right) = 0\\
\to \left[ \begin{array}{l}
x - 1 = 0\\
2x + 5 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1\\
x = - \dfrac{5}{2}
\end{array} \right.\\
e)2x\left( {3x - 2} \right) - \left( {3x - 1} \right)\left( {3x - 2} \right) = 0\\
\to \left( {3x - 2} \right)\left( {2x - 3x + 1} \right) = 0\\
\to \left[ \begin{array}{l}
3x - 2 = 0\\
- x + 1 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{2}{3}\\
x = 1
\end{array} \right.\\
b)3\left( {x - 5} \right)\left( {x + 2} \right) - x\left( {x - 5} \right) = 0\\
\to \left( {x - 5} \right)\left( {3x + 6 - x} \right) = 0\\
\to \left[ \begin{array}{l}
x - 5 = 0\\
2x + 6 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 5\\
x = - 3
\end{array} \right.\\
d) - \dfrac{{x - 7}}{2} + \dfrac{{2\left( {x - 7} \right)\left( {x - 3} \right)}}{3} = 0\\
\to \left( {x - 7} \right)\left( { - \dfrac{1}{2} + \dfrac{2}{3}x - 2} \right) = 0\\
\to \left[ \begin{array}{l}
x - 7 = 0\\
\dfrac{2}{3}x - \dfrac{5}{2} = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 7\\
x = \dfrac{{15}}{4}
\end{array} \right.\\
f){\left( {2x - 1} \right)^2} + \left( {x - 3} \right)\left( {2x - 1} \right) = 0\\
\to \left( {2x - 1} \right)\left( {2x - 1 + x - 3} \right) = 0\\
\to \left[ \begin{array}{l}
2x - 1 = 0\\
3x - 4 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x = \dfrac{4}{3}
\end{array} \right.
\end{array}\)
