$a$) $|4x^2 - 25| = 0$
$⇔ 4x^2 - 25 = 0$
$⇔ 4x^2 = 25$
$⇔ x^2 = \dfrac{25}{4}$
$⇔ x = ±\dfrac{5}{2}$
Vậy $ x = ±\dfrac{5}{2}$
$b$) $|x-2|=3$
$⇒$ \(\left[ \begin{array}{l}x-2=3\\x-2=-3\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}x=5\\x=-1\end{array} \right.\)
Vậy $x$ $∈$ `{5;-1}`
$c$) $|x-3| = 2x-1$ ($1$)
$TH1$.$x-3$ $≥$ $0$ $⇔$ $x$ $≥$ $3$
Từ ($1$) $⇒$ $x-3 = 2x-1$
$⇔ -3 + 1 = 2x - x$
$⇔ x = -2$ ($KTM$)
$TH2$.$x-3$ $<$ $0$ $⇔$ $x$ $<$ $3$
Từ ($1$) $⇒$ $-x+3 = 2x-1$
$⇔ 3 + 1 = 2x + x$
$⇔ x = \dfrac{4}{3}$ ($KTM$)
Vậy $S$=`∅`
$d$) $|3x-2|=x+5$ ($2$)
$TH1$.$3x-2$ $≥$ $0$ $⇔$ $x$ $≥$ $\dfrac{2}{3}$
Từ ($2$) $⇒$ $3x-2 = x+5$
$⇔ 3x - x = 5 + 2$
$ ⇔ 2x = 7$
$⇔ x = \dfrac{7}{2}$ ($TM$)
$TH2$.$3x-2$ $<$ $0$ $⇔$ $x$ $<$ $\dfrac{2}{3}$
Từ ($2$) $⇒$ $-3x+2 = x+5$
$⇔ 2-5 = x + 3x$
$ ⇔ 4x = -3$
$⇔ x = \dfrac{-3}{4}$ ($TM$)
Vậy $S$=`{\frac{7}{2};\frac{-3}{4}}`
