`|2x – 1| – |x + 1/3| = 0` `<=> |2x – 1| = |x + 1/3|` `<=>` \(\left[ \begin{array}{l}2x – 1 = x + \dfrac{1}{3}\\2x – 1 = -x – \dfrac{1}{3}\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x = \dfrac{4}{3}\\x = \dfrac{2}{9}\end{array} \right.\) Reply
Đáp án: Giải thích các bước giải: `|2x-1|-|x+1/3|=0``=>|2x-1|=|x+1/3|`TH1`2x-1=x+1/3``=>x=1/3+1=4/3`TH2`2x-1=-x-1/3``=>3x=2/3``=>x=2/9` Reply
`|2x – 1| – |x + 1/3| = 0`
`<=> |2x – 1| = |x + 1/3|`
`<=>` \(\left[ \begin{array}{l}2x – 1 = x + \dfrac{1}{3}\\2x – 1 = -x – \dfrac{1}{3}\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x = \dfrac{4}{3}\\x = \dfrac{2}{9}\end{array} \right.\)
Đáp án:
Giải thích các bước giải:
`|2x-1|-|x+1/3|=0`
`=>|2x-1|=|x+1/3|`
TH1
`2x-1=x+1/3`
`=>x=1/3+1=4/3`
TH2
`2x-1=-x-1/3`
`=>3x=2/3`
`=>x=2/9`