Đáp án:
k) \(\left[ \begin{array}{l}
x = \dfrac{8}{3}\\
x = - 2
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)A = 0\\
\to - 2x + \dfrac{1}{3} = 0\\
\to x = \dfrac{1}{6}\\
b)B = 0\\
\to 3\left( {2x - 1} \right) - 2 = 0\\
\to 2x - 1 = \dfrac{2}{3}\\
\to 2x = \dfrac{5}{3}\\
\to x = \dfrac{5}{6}\\
c)2\left( {x - 1} \right) - 3\left( {x - 2} \right) = 0\\
\to 2x - 2 - 3x + 6 = 0\\
\to - x + 4 = 0\\
\to x = 4\\
d) - \dfrac{1}{2}x - \dfrac{2}{5}\left( {x + 1} \right) = 0\\
\to - \dfrac{1}{2}x - \dfrac{2}{5}x - \dfrac{2}{5} = 0\\
\to - \dfrac{9}{{10}}x = \dfrac{2}{5}\\
\to x = - \dfrac{4}{9}\\
e)x\left( {x + 5} \right)\left( {x - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x + 5 = 0\\
x - 1 = 0
\end{array} \right. \to \left[ \begin{array}{l}
x = 0\\
x = - 5\\
x = 1
\end{array} \right.\\
f)5\left( {\dfrac{x}{2} - 5} \right)\left( {x + \dfrac{1}{4}} \right) = 0\\
\to \left[ \begin{array}{l}
\dfrac{x}{2} - 5 = 0\\
x + \dfrac{1}{4} = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 10\\
x = - \dfrac{1}{4}
\end{array} \right.\\
g)2\left( {x - 1} \right) - 5\left( {x + 2} \right) + 10 = 0\\
\to 2x - 2 - 5x - 10 + 10 = 0\\
\to - 3x = 2\\
\to x = - \dfrac{2}{3}\\
h)9{x^2} - 16 = 0\\
\to 9{x^2} = 16\\
\to {x^2} = \dfrac{{16}}{9}\\
\to \left[ \begin{array}{l}
x = \dfrac{4}{3}\\
x = - \dfrac{4}{3}
\end{array} \right.\\
i)\left| {2x - 3} \right| - 11 = 0\\
\to \left| {2x - 3} \right| = 11\\
\to \left[ \begin{array}{l}
2x - 3 = 11\\
2x - 3 = - 11
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 7\\
x = - 4
\end{array} \right.\\
k)\left| {x - 5} \right| = 3 - 2x\\
\to \left[ \begin{array}{l}
x - 5 = 3 - 2x\\
x - 5 = - 3 + 2x
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{8}{3}\\
x = - 2
\end{array} \right.
\end{array}\)
