`| 5/2 – x | = 13/10` `| x + 3/4 | + 1/3 = 0` ` | x – 3/2 | + | 5/2 – x | = 0` 30/07/2021 Bởi Melanie `| 5/2 – x | = 13/10` `| x + 3/4 | + 1/3 = 0` ` | x – 3/2 | + | 5/2 – x | = 0`
`|5/2 – x| = 13/10` `⇒` \(\left[ \begin{array}{l}\dfrac{5}{2}-x=\dfrac{13}{10}\\\dfrac{5}{2}-x=-\dfrac{13}{10}\end{array} \right.\) $⇒$ \(\left[ \begin{array}{l}x= \dfrac{6}{5}\\x = \dfrac{19}{5}\end{array} \right.\) Vậy `x` `∈` `{19/5;6/5}` `|x+3/4| + 1/3 = 0` `⇔ |x+3/4| = -1/3` Vì : `|x+3/4| ≥ 0 ∀ x` `⇒ x ∈`∅ Vậy `x` `∈` ∅ `|x-3/2| + |5/2 – x| = 0` Vì : `|x-3/2|;|5/2-x| ≥ 0 ∀ x` `⇒` `x-3/2 = 5/2 – x = 0` `⇒` `x=3/2;x = 5/2` (Không thỏa mãn) Vậy `x` `∈` $\varnothing$. Bình luận
`|5/2 – x| = 13/10`
`⇒` \(\left[ \begin{array}{l}\dfrac{5}{2}-x=\dfrac{13}{10}\\\dfrac{5}{2}-x=-\dfrac{13}{10}\end{array} \right.\)
$⇒$ \(\left[ \begin{array}{l}x= \dfrac{6}{5}\\x = \dfrac{19}{5}\end{array} \right.\)
Vậy `x` `∈` `{19/5;6/5}`
`|x+3/4| + 1/3 = 0`
`⇔ |x+3/4| = -1/3`
Vì : `|x+3/4| ≥ 0 ∀ x`
`⇒ x ∈`∅
Vậy `x` `∈` ∅
`|x-3/2| + |5/2 – x| = 0`
Vì : `|x-3/2|;|5/2-x| ≥ 0 ∀ x`
`⇒` `x-3/2 = 5/2 – x = 0`
`⇒` `x=3/2;x = 5/2` (Không thỏa mãn)
Vậy `x` `∈` $\varnothing$.