`d, |2x - 1| + |1 - 3x| = 2x + 1`
$\text{TH1: Có: |2x - 1| = 2x - 1 nếu 2x - 1 ≥ 0 ⇔ x ≥ $\dfrac{1}{2}$}$
$\text{|1 - 3x| = 1 - 3x nếu 1 - 3x ≥ 0 ⇔ x ≤ $\dfrac{1}{3}$}$
`⇒ x ∈ ∅`
$\text{TH2: Có: |2x - 1| = -(2x - 1) = 1 - 2x nếu 2x - 1 < 0 ⇔ x < $\dfrac{1}{2}$}$
$\text{|1 - 3x| = -(1 - 3x) = 3x - 1 nếu 1 - 3x < 0 ⇔ x > $\dfrac{1}{3}$}$
`⇒ 1/2 > x > 1/3`
$\text{Với $\dfrac{1}{2}$ > x > $\dfrac{1}{3}$}$
`⇒ 1 - 2x + 3x - 1 = 2x + 1`
`⇔ -x = 0`
`⇔ x = 0 (KTM)`
$\text{TH3: Có: |2x - 1| = 2x - 1 nếu 2x - 1 ≥ 0 ⇔ x ≥ $\dfrac{1}{2}$}$
$\text{|1 - 3x| = -(1 - 3x) = 3x - 1 nếu 1 - 3x < 0 ⇔ x > $\dfrac{1}{3}$}$
`⇒ x ≥ 1/2`
$\text{Với x ≥ $\dfrac{1}{2}$}$
`⇒ 2x - 1 + 3x - 1 = 2x + 1`
`⇔ 3x = 3`
`⇔ x = 1 (TM)`
$\text{TH4: Có: |2x - 1| = -(2x - 1) = 1 - 2x nếu 2x - 1 < 0 ⇔ x < $\dfrac{1}{2}$}$
$\text{|1 - 3x| = 1 - 3x nếu 1 - 3x ≥ 0 ⇔ x ≤ $\dfrac{1}{3}$}$
`⇒ x ≤ 1/3`
$\text{Với x ≤ $\dfrac{1}{3}$}$
`⇒ 1 - 2x + 1 - 3x = 2x + 1`
`⇔ -7x = -1`
`⇔ x = 1/7 (TM)`
$\text{Vậy x = 1; x = $\dfrac{1}{7}$}$
`e, |x + 1| - |2x + 3| = (-x) + 2`
$\text{TH1: Có: |x + 1| = x + 1 nếu x + 1 ≥ 0 ⇔ x ≥ -1}$
$\text{|2x + 3| = 2x + 3 nếu 2x + 3 ≥ 0 ⇔ x ≥ $\dfrac{-3}{2}$}$
`⇒ x ≥ -1`
$\text{Với x ≥ -1}$
`⇒ x + 1 - 2x - 3 = -x + 2`
$\text{⇔ 0 = 4 (vô lí)}$
`⇒ x ∈ ∅`
$\text{TH2: Có: |x + 1| = -x - 1 nếu x + 1 < 0 ⇔ x < -1}$
$\text{|2x + 3| = -2x - 3 nếu 2x + 3 < 0 ⇔ x < $\dfrac{-3}{2}$}$
`⇒ x < -3/2`
$\text{Với x < $\dfrac{-3}{2}$}$
`⇒ -x - 1 - (-2x - 3) = -x + 2`
`⇔ -x - 1 + 2x + 3 = -x + 2`
`⇔ 2x = 0`
`⇔ x = 0 (KTM)`
$\text{TH3: Có: |x + 1| = x + 1 nếu x + 1 ≥ 0 ⇔ x ≥ -1}$
$\text{|2x + 3| = -2x - 3 nếu 2x + 3 < 0 ⇔ x < $\dfrac{-3}{2}$}$
`⇒ x ∈ ∅`
$\text{TH4: Có: |x + 1| = -x - 1 nếu x + 1 < 0 ⇔ x < -1}$
$\text{|2x + 3| = 2x + 3 nếu 2x + 3 ≥ 0 ⇔ x ≥ $\dfrac{-3}{2}$}$
`⇒ -3/2 ≤ x < -1`
$\text{Với $\dfrac{-3}{2}$ ≤ x < -1}$
`⇒ -x - 1 - (2x + 3) = -x + 2`
`⇔ -x - 1 - 2x - 3 = -x + 2`
`⇔ -2x = 6`
`⇔ x = -3 (KTM)`
$\text{Vậy S = ∅}$
