Đáp án:
\(\begin{array}{l}
a)\left[ \begin{array}{l}
x = - \dfrac{1}{4}\\
x = - \dfrac{5}{4}
\end{array} \right.\\
b)\left[ \begin{array}{l}
x = \dfrac{{63}}{{16}}\\
x = - \dfrac{{63}}{{16}}
\end{array} \right.\\
c)\left[ \begin{array}{l}
x = \dfrac{3}{2}\\
x = \dfrac{5}{2}
\end{array} \right.\\
d)\left[ \begin{array}{l}
x = \dfrac{{13}}{{20}}\\
x = \dfrac{3}{{20}}
\end{array} \right.\\
e)\left[ \begin{array}{l}
x = - \dfrac{4}{3}\\
x = - 2
\end{array} \right.\\
f)x = \dfrac{1}{5}\\
g)x \in \emptyset
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\left| {x + \dfrac{3}{4}} \right| = \dfrac{1}{2}\\
\to \left[ \begin{array}{l}
x + \dfrac{3}{4} = \dfrac{1}{2}\\
x + \dfrac{3}{4} = - \dfrac{1}{2}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - \dfrac{1}{4}\\
x = - \dfrac{5}{4}
\end{array} \right.\\
b)\left| {4x} \right| - \dfrac{{27}}{2} = \dfrac{9}{4}\\
\to \left| {4x} \right| = \dfrac{{63}}{4}\\
\to \left[ \begin{array}{l}
4x = \dfrac{{63}}{4}\\
4x = - \dfrac{{63}}{4}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{63}}{{16}}\\
x = - \dfrac{{63}}{{16}}
\end{array} \right.\\
c)\left| {2 - x} \right| = \dfrac{5}{6} - \dfrac{1}{3}\\
\to \left| {2 - x} \right| = \dfrac{1}{2}\\
\to \left[ \begin{array}{l}
2 - x = \dfrac{1}{2}\\
2 - x = - \dfrac{1}{2}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{3}{2}\\
x = \dfrac{5}{2}
\end{array} \right.\\
d)\left| {x - \dfrac{2}{5}} \right| = \dfrac{1}{4}\\
\to \left[ \begin{array}{l}
x - \dfrac{2}{5} = \dfrac{1}{4}\\
x - \dfrac{2}{5} = - \dfrac{1}{4}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{13}}{{20}}\\
x = \dfrac{3}{{20}}
\end{array} \right.\\
e)\left| {3x + 5} \right| = 1\\
\to \left[ \begin{array}{l}
3x + 5 = 1\\
3x + 5 = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - \dfrac{4}{3}\\
x = - 2
\end{array} \right.\\
f)\left| {\dfrac{1}{5} - x} \right| = 0\\
\to x = \dfrac{1}{5}\\
g)\left| {5 - 3x} \right| = - \dfrac{1}{2}\\
Do:\left| {5 - 3x} \right| \ge 0\forall x\\
\to \left| {5 - 3x} \right| = - \dfrac{1}{2}\left( {KTM} \right)\\
\to x \in \emptyset
\end{array}\)
