|3x-1| +4=6-|3x-1|
⇔ |3x-1| +|3x-1| =-4+6
⇔ 2|3x-1| = 2
⇔ |3x-1|=1
ta có
|3x-1|=3x-1 thì 3x-1 ≥ 0 ⇔ x ≥ \(\dfrac{1}{3}\)
|3x-1=-(3x-1) thì 3x-1 < 0 ⇔ x< \(\dfrac{1}{3}\)
th1 với x≥\(\dfrac{1}{3}\)
3x-1=1
⇔ 3x=2
⇔ x=\(\dfrac{2}{3}\) (tm)
th2 với x < \(\dfrac{1}{3}\)
-(3x-1)=1
⇔ -3x+1=1
⇔ -3x=0
⇔ x=0 (tm)
vậy pt có tập nghiệm S=\(\left\{0;\dfrac{2}{3}\right\}\)
chắc vậy
\(\left|3x-1\right|+4=6-\left|3x-1\right|\left(1\right)\)
t/h1: \(3x-1\ge0\Leftrightarrow x\ge\dfrac{1}{3}\)
(1) \(\Leftrightarrow3x-1+4=6-3x+1\)
\(\Leftrightarrow6x=4\)
\(\Leftrightarrow x=\dfrac{2}{3}\) TĐK
t/h 2: \(3x-1< 0\Leftrightarrow x< \dfrac{1}{3}\)
(1) \(\Leftrightarrow-3x+1+4=6+3x-1\)
\(\Leftrightarrow-6x=0\)
\(\Leftrightarrow x=0\) TĐK
\(\Rightarrow S=\left\{0;\dfrac{2}{3}\right\}\)
