Đáp án:
$a)| 2x-3| =4$
Th1 :
$2x-3 =4$
$⇔2x = 7$
$⇔x=\dfrac{7}{2}$
Th2 :
$2x-3 = -4$
$⇔2x =-1$
$⇔x=-\dfrac{1}{2}$
$\text{Vậy S={$\dfrac{7}{2} ; -\dfrac{1}{2}$}}$
$b) |3x -1| -x =2$
$⇔|3x-1| =2+x$
Th1 : $3x-1 ≥ 0 ⇔ x ≥ \dfrac{1}{3}$
$⇔3x-1 =2+x$
$⇔2x=3$
$⇔x=\dfrac{3}{2}(TM)$
Th2 :$3x-1 ≤ 0 ⇔ x ≤ \dfrac{1}{3}$
$⇔3x-1 =-2-x$
$⇔4x =-1$
$⇔x=-\dfrac{1}{4}(TM)$
$\text{Vậy S={$\dfrac{3}{2} ; -\dfrac{1}{4}$}}$
$c) |x-7|=2x+3$
Th1 : $x -7 ≥ 0 ⇔ x ≥ 7$
$⇔x-7 =2x+3$
$⇔-x= 10$
$⇔x=-10 (KTM)$
TH2 : $x-7 ≤ 0 ⇔ x ≤ 7$
$⇔x-7 = -2x -3$
$⇔3x=4$
$⇔x=\dfrac{4}{3} (TM)$
$\text{Vậy S={$\dfrac{4}{3}$}}$
$d) |x-4| +3x=5$
$⇔|x-4| = 5-3x$
Th1 :$x-4 ≥ 0 ⇔ x ≥ 4$
$⇔x-4 =5-3x$
$⇔4x =9$
$⇔x=\dfrac{9}{4}(KTM)$
Th2 : $x-4 ≤ 0 ⇔ x ≤ 4$
$⇔x-4= -5+3x$
$⇔-2x=-1$
$⇔x=\dfrac{1}{2}(TM)$
$\text{Vậy S={$\dfrac{1}{2}$}}$
$e) |x-7| =2$
Th1 : $x-7 =2$
$⇔x=9$
Th2 : $x-7=-2$
$⇔x=5$
$\text{Vậy S={9 ; 5} }$
$g) |5-2x| -1=x$
$⇔|5-2x| =x+1$
Th1 : $5-2x ≥ 0⇔x ≤ \dfrac{5}{2}$
$⇔5-2x= x +1$
$⇔-3x =-4$
$⇔x=\dfrac{4}{3}(TM)$
Th2 : $5-2x ≤ 0 ⇔ x ≥ \dfrac{5}{2}$
$⇔5-2x =-x-1$
$⇔-x=-6$
$⇔x=6 (TM)$
$\text{Vậy S={$\dfrac{4}{3}$ ; 6$}}$
