`a) | 1 - 2x | - 5x = x + 2`

`⇒ | 1 - 2x | = x + 2 + 5x`

`⇒ | 1 - 2x | = 6x + 2`

Điều kiện `: x ≥ ( - 1 )/3`

Khi đó `: | 1 - 2x | = 6x + 2`

`⇔` \(\left[ \begin{array}{l}1 - 2x = 6x + 2\\1 - 2x = - 6x - 2\end{array} \right.\) 

`⇔` \(\left[ \begin{array}{l}2x + 6x = 1 + 2\\- 6x + 2x = 1 + 2\end{array} \right.\) 

`⇔` \(\left[ \begin{array}{l}8x = 3\\- 4x = 3\end{array} \right.\) 

`⇔` \(\left[ \begin{array}{l}x = \frac{3}{8} ( t/m ) \\x = \frac{- 3}{4} ( loại ) \end{array} \right.\) 

Vậy `,` tập nghiệm `S = { 3/8 } .`

`b) | x + 5 | = 5x + 1 ( ĐK : x ≥ ( - 1 )/5 )`

`⇔` \(\left[ \begin{array}{l}x + 5 = 5x + 1\\x + 5 = - 5x - 1\end{array} \right.\) 

`⇔` \(\left[ \begin{array}{l}5x - x = 5 - 1\\ - 5x - x = 1 + 5\end{array} \right.\) 

`⇔` \(\left[ \begin{array}{l}4x = 4\\- 6x = 6\end{array} \right.\) 

`⇔` \(\left[ \begin{array}{l}x = 1 ( t/m ) \\x = - 1 ( Loại )\end{array} \right.\) 

Vậy `,` tập nghiệm `S = { 1 } .`

`Parkchaeyoung`

`a, |1-2x|-5x=x+2`

`<=> |1-2x|=x+2+5x`

`<=> |1-2x|=6x+2` `(1)`

`TH1: 1-2x>=0=>x<=1/2=>|1-2x|=1-2x`

`(1)=> 1-2x=6x+2`

`<=> -2x-6x=2-1`

`<=> -8x=1`

`<=> x=-1/8` `(tm)`

`TH2: 1-2x<0=>x>1/2=>|1-2x|=-1+2x`

`(1)=> -1+2x=6x+2`

`<=> 2x-6x=2+1`

`<=> -4x=3`

`<=> x=-3/4` (ko tm)

Vậy: ...........

`b, |x+5|=5x+1` `(2)`

`TH1: x+5>=0=>x>=-5=>|x+5|=x+5`

`(2)=> x+5=5x+1`

`<=> x-5x=1-5`

`<=> -4x=-4`

`<=> x=1` `(tm)`

`TH2: x+5<0=>x<-5=>|x+5|=-x-5`

`(2)=> -x-5=5x+1`

`<=> -x-5x=1+5`

`<=> -6x=6`

`<=> x=-1` (ko tm)

Vậy: ......................