d) |2x-1|+|1-3x|= 2x+1 e) |x+1|- |2x+3| = (-x) + 2

0 bình luận về “d) |2x-1|+|1-3x|= 2x+1 e) |x+1|- |2x+3| = (-x) + 2”

1. Đáp án:

   Câu e bạn cũng làm 4 trường hợp giống mk nhé xong bạn tự giải ra là đc r???? 

   Giải thích các bước giải:

    

   Trả lời

2. `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)`

   $\text{Vậy S = ∅}$

   Trả lời