$\textit{Đáp án + Giải thích các bước giải:}$
$\text{a) 6 - $\bigg|\dfrac{1}{2} - x\bigg|$ = $\dfrac{2}{5}$}$
$\text{$\bigg|\dfrac{1}{2} - x\bigg|$ = 6 - $\dfrac{2}{5}$ = $\dfrac{28}{5}$}$
$\text{$\dfrac{1}{2}$ - $x^{}$ = ± $\dfrac{28}{5}$}$
\(\left[ \begin{array}{l}\dfrac{1}{2} - x = \dfrac{28}{5}\\\dfrac{1}{2} - x = \dfrac{-28}{5}\end{array} \right.\)
\(\left[ \begin{array}{l}x = \dfrac{1}{2} - \dfrac{28}{5}\\x=\dfrac{1}{2} - \dfrac{-28}{5}\end{array} \right.\)\(\left[ \begin{array}{l}x=\dfrac{5}{10} - \dfrac{56}{10} = \dfrac{-41}{10}\\x=\dfrac{5}{10} + \dfrac{56}{10} = \dfrac{61}{10}\end{array} \right.\)
$\text{Vậy $x^{}$ ∈ {$\dfrac{61}{10}$; $\dfrac{-41}{10}$}}$
$\text{b) $\bigg|x + \dfrac{3}{5}\bigg|$ - $\dfrac{1}{2}$ = $\dfrac{1}{2}$}$
$\text{$\bigg|x + \dfrac{3}{5}\bigg|$ = $\dfrac{1}{2}$ + $\dfrac{1}{2}$ = 1}$
$\text{$x^{}$ + $\dfrac{3}{5}$ = ± 1}$
$\text{\(\left[ \begin{array}{l}x + \dfrac{3}{5} = 1\\x + \dfrac{3}{5} = -1\end{array} \right.\) }$
\(\left[ \begin{array}{l}x=1 - \dfrac{3}{5} = \dfrac{2}{5}\\x=-1 - \dfrac{3}{5} = \dfrac{-8}{5}\end{array} \right.\)
$\text{Vậy $x^{}$ ∈ {$\dfrac{2}{5}$; $\dfrac{-8}{5}$}}$
