Toán d) |2x-1|+|1-3x|= 2x+1 e) |x+1|- |2x+3| = (-x) + 2 16/09/2021 By Charlie d) |2x-1|+|1-3x|= 2x+1 e) |x+1|- |2x+3| = (-x) + 2
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
`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
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:
`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)`
$\text{Vậy S = ∅}$