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

Question

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

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Charlie 3 tháng 2021-09-16T14:16:52+00:00 2 Answers 11 views 0

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    0
    2021-09-16T14:18:43+00:00

    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:

     

    0
    2021-09-16T14:18:43+00:00

    `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)`

    $\text{Vậy S = ∅}$

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