Đáp án:
Bài 1 :
$a) |x| =\dfrac{2}{3}$
$⇔x =±\dfrac{2}{3}$
$b |x+\dfrac{1}{2}| = 0,3$
⇔\(\left[ \begin{array}{l}x+\dfrac{1}{2}=0,3\\x+\dfrac{1}{2}=-0,3\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-\dfrac{1}{5}\\x=-\dfrac{4}{5}\end{array} \right.\)
Vậy $\text{x ∈ { -$\dfrac{1}{5}$ ; -$\dfrac{4}{5}$}}$
$c) | \dfrac{2}{5} - x | = 1$
⇔\(\left[ \begin{array}{l}\dfrac{2}{5}-x=1\\\dfrac{2}{5}-x=-1\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-\dfrac{3}{5}\\x=\dfrac{7}{5}\end{array} \right.\)
Vậy $\text{x ∈ { $-\dfrac{3}{5} ; \dfrac{7}{5}$}}$
$d) |x +2| -\dfrac{1}{2} = 0$
$⇔|x+2 | = \dfrac{1}{2}$
⇔\(\left[ \begin{array}{l}x+2=\dfrac{1}{2}\\x+2=-\dfrac{1}{2}\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=-\dfrac{3}{2}\\x=-\dfrac{5}{2}\end{array} \right.\)
Vậy $\text{x ∈ {$-\dfrac{3}{2} ; -\dfrac{5}{2}$}}$
$) 1,6 - |x - 0,2| = 0$
$⇔-|x-0,2| = -1,6$
$⇔|x-0,2| = 1,6$
⇔\(\left[ \begin{array}{l}x-0,2=1,6\\x-0,2=-1,6\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=1,8\\x=-1,4\end{array} \right.\)
Vậy $\text{x ∈ { $1,8 ; -1,4$}}$
Bài 2 :
$A = 0,5 - |x-3,5|$
Vì $-|x-3,5| ≤ 0$
Nên $|x-3,5| +0,5 ≤ 0,5$
Dấu ''='' xảy ra khi $x-3,5 = 0 ⇔x= 3,5$
Vậy Max A $=0,5$ , tại $x = 3,5$
