Đáp án:
$\begin{array}{l}
2)\left| {3x} \right| = \dfrac{1}{3} - \dfrac{1}{4}\\
\Rightarrow \left| {3x} \right| = \dfrac{1}{{12}}\\
\Rightarrow \left[ \begin{array}{l}
3x = \dfrac{1}{{12}}\\
3x = - \dfrac{1}{{12}}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{1}{{36}}\\
x = - \dfrac{1}{{36}}
\end{array} \right.\\
Vậy\,x = \dfrac{1}{{36}};x = - \dfrac{1}{{36}}\\
3)\left| {2x} \right| = \dfrac{1}{4}\\
\Rightarrow \left[ \begin{array}{l}
2x = \dfrac{1}{4}\\
2x = - \dfrac{1}{4}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{1}{8}\\
x = - \dfrac{1}{8}
\end{array} \right.\\
Vậy\,x = \dfrac{1}{8};x = - \dfrac{1}{8}\\
6)\left| { - x - 1} \right| = \dfrac{1}{6}\\
\Rightarrow \left[ \begin{array}{l}
- x - 1 = \dfrac{1}{6}\\
- x - 1 = - \dfrac{1}{6}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = - 1 - \dfrac{1}{6} = - \dfrac{7}{6}\\
x = - 1 + \dfrac{1}{6} = - \dfrac{5}{6}
\end{array} \right.\\
Vậy\,x = - \dfrac{7}{6};x = - \dfrac{5}{6}\\
9)\left| {x - 4} \right| = \dfrac{1}{3} - \dfrac{1}{7}\\
\Rightarrow \left| {x - 4} \right| = \dfrac{4}{{21}}\\
\Rightarrow \left[ \begin{array}{l}
x - 4 = \dfrac{4}{{21}}\\
x - 4 = - \dfrac{4}{{21}}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 4 + \dfrac{4}{{21}} = \dfrac{{88}}{{21}}\\
x = 4 - \dfrac{4}{{21}} = \dfrac{{80}}{{21}}
\end{array} \right.\\
10)\left| {x - \dfrac{1}{4}} \right| = \dfrac{1}{7} - \dfrac{1}{3}\\
\Rightarrow \left| {x - \dfrac{1}{4}} \right| = - \dfrac{4}{{21}}\left( {ktm} \right)\\
Do:2)\left| {3x} \right| = \dfrac{1}{3} - \dfrac{1}{4}\\
\Rightarrow \left| {3x} \right| = \dfrac{1}{{12}}\\
\Rightarrow \left[ \begin{array}{l}
3x = \dfrac{1}{{12}}\\
3x = - \dfrac{1}{{12}}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{1}{{36}}\\
x = - \dfrac{1}{{36}}
\end{array} \right.\\
Vậy\,x = \dfrac{1}{{36}};x = - \dfrac{1}{{36}}\\
3)\left| {2x} \right| = \dfrac{1}{4}\\
\Rightarrow \left[ \begin{array}{l}
2x = \dfrac{1}{4}\\
2x = - \dfrac{1}{4}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = \dfrac{1}{8}\\
x = - \dfrac{1}{8}
\end{array} \right.\\
Vậy\,x = \dfrac{1}{8};x = - \dfrac{1}{8}\\
6)\left| { - x - 1} \right| = \dfrac{1}{6}\\
\Rightarrow \left[ \begin{array}{l}
- x - 1 = \dfrac{1}{6}\\
- x - 1 = - \dfrac{1}{6}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = - 1 - \dfrac{1}{6} = - \dfrac{7}{6}\\
x = - 1 + \dfrac{1}{6} = - \dfrac{5}{6}
\end{array} \right.\\
Vậy\,x = - \dfrac{7}{6};x = - \dfrac{5}{6}\\
9)\left| {x - 4} \right| = \dfrac{1}{3} - \dfrac{1}{7}\\
\Rightarrow \left| {x - 4} \right| = \dfrac{4}{{21}}\\
\Rightarrow \left[ \begin{array}{l}
x - 4 = \dfrac{4}{{21}}\\
x - 4 = - \dfrac{4}{{21}}
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 4 + \dfrac{4}{{21}} = \dfrac{{88}}{{21}}\\
x = 4 - \dfrac{4}{{21}} = \dfrac{{80}}{{21}}
\end{array} \right.\\
10)\left| {x - \dfrac{1}{4}} \right| = \dfrac{1}{7} - \dfrac{1}{3}\\
\Rightarrow \left| {x - \dfrac{1}{4}} \right| = - \dfrac{4}{{21}}\left( {ktm} \right)
\end{array}$
Do trị tuyệt đối luôn lớn hơn hoặc bằng 0 nên pt vô nghiệm
