Đáp án:
\(\begin{array}{l}
4,\\
a,\,\,\,\,\left[ \begin{array}{l}
x = \dfrac{4}{3}\\
x = - \dfrac{{28}}{{15}}
\end{array} \right.\\
b,\,\,\,\,\left[ \begin{array}{l}
x = 1\\
x = - \dfrac{1}{3}
\end{array} \right.\\
5,\\
a,\,\,\,\,\left[ \begin{array}{l}
x = - \dfrac{{13}}{5}\\
x = \dfrac{{17}}{5}
\end{array} \right.\\
b,\,\,\,\,\left[ \begin{array}{l}
x = \dfrac{{38}}{{15}}\\
x = - \dfrac{{28}}{{15}}
\end{array} \right.\\
c,\,\,\,\,\left[ \begin{array}{l}
x = \dfrac{2}{3}\\
x = - \dfrac{1}{3}
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
4,\\
a,\\
\left| {x + \dfrac{4}{{15}}} \right| - \left| { - 3,75} \right| = - \left| { - 2,15} \right|\\
\Leftrightarrow \left| {x + \dfrac{4}{{15}}} \right| - 3,75 = - 2,15\\
\Leftrightarrow \left| {x + \dfrac{4}{{15}}} \right| = - 2,15 + 3,75\\
\Leftrightarrow \left| {x + \dfrac{4}{{15}}} \right| = 1,6\\
\Leftrightarrow \left| {x + \dfrac{4}{{15}}} \right| = \dfrac{8}{5}\\
\Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{4}{{15}} = \dfrac{8}{5}\\
x + \dfrac{4}{{15}} = - \dfrac{8}{5}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{8}{5} - \dfrac{4}{{15}}\\
x = - \dfrac{8}{5} - \dfrac{4}{{15}}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{4}{3}\\
x = - \dfrac{{28}}{{15}}
\end{array} \right.\\
b,\\
2\left| {3x - 1} \right| + 1 = 5\\
\Leftrightarrow 2\left| {3x - 1} \right| = 5 - 1\\
\Leftrightarrow 2\left| {3x - 1} \right| = 4\\
\Leftrightarrow \left| {3x - 1} \right| = 4:2\\
\Leftrightarrow \left| {3x - 1} \right| = 2\\
\Leftrightarrow \left[ \begin{array}{l}
3x - 1 = 2\\
3x - 1 = - 2
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
3x = 2 + 1\\
3x = - 2 + 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
3x = 3\\
3x = - 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = - \dfrac{1}{3}
\end{array} \right.\\
5,\\
a,\\
\left| { - x + \dfrac{2}{5}} \right| + \dfrac{1}{2} = 3,5\\
\Leftrightarrow \left| { - x + \dfrac{2}{5}} \right| + \dfrac{1}{2} = \dfrac{7}{2}\\
\Leftrightarrow \left| { - x + \dfrac{2}{5}} \right| = \dfrac{7}{2} - \dfrac{1}{2}\\
\Leftrightarrow \left| { - x + \dfrac{2}{5}} \right| = 3\\
\Leftrightarrow \left[ \begin{array}{l}
- x + \dfrac{2}{5} = 3\\
- x + \dfrac{2}{5} = - 3
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
- x = 3 - \dfrac{2}{5}\\
- x = - 3 - \dfrac{2}{5}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
- x = \dfrac{{13}}{5}\\
- x = - \dfrac{{17}}{5}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = - \dfrac{{13}}{5}\\
x = \dfrac{{17}}{5}
\end{array} \right.\\
b,\\
\left| {x - \dfrac{1}{3}} \right| = 2\dfrac{1}{5}\\
\Leftrightarrow \left| {x - \dfrac{1}{3}} \right| = \dfrac{{11}}{5}\\
\Leftrightarrow \left[ \begin{array}{l}
x - \dfrac{1}{3} = \dfrac{{11}}{5}\\
x - \dfrac{1}{3} = - \dfrac{{11}}{5}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{11}}{5} + \dfrac{1}{3}\\
x = - \dfrac{{11}}{5} + \dfrac{1}{3}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{38}}{{15}}\\
x = - \dfrac{{28}}{{15}}
\end{array} \right.\\
c,\\
2 - \left| {\dfrac{3}{2}x - \dfrac{1}{4}} \right| = \left| { - \dfrac{5}{4}} \right|\\
\Leftrightarrow 2 - \left| {\dfrac{3}{2}x - \dfrac{1}{4}} \right| = \dfrac{5}{4}\\
\Leftrightarrow \left| {\dfrac{3}{2}x - \dfrac{1}{4}} \right| = 2 - \dfrac{5}{4}\\
\Leftrightarrow \left| {\dfrac{3}{2}x - \dfrac{1}{4}} \right| = \dfrac{3}{4}\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{3}{2}x - \dfrac{1}{4} = \dfrac{3}{4}\\
\dfrac{3}{2}x - \dfrac{1}{4} = - \dfrac{3}{4}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\dfrac{3}{2}x = \dfrac{3}{4} + \dfrac{1}{4}\\
\dfrac{3}{2}x = - \dfrac{3}{4} + \dfrac{1}{4}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{3}{2}x = 1\\
\dfrac{3}{2}x = - \dfrac{1}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1:\dfrac{3}{2}\\
x = - \dfrac{1}{2}:\dfrac{3}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{2}{3}\\
x = - \dfrac{1}{3}
\end{array} \right.
\end{array}\)
