a) | x + $\frac{4}{15}$ | – | – 3,75 | = – | – 2, 15 | b) | 5x – 4 | = | x + 2 | c) [| 2x – 1 | + $\frac{1}{2}$ | = $\frac{4}{5}$ d) | x + 4 | + | 9

Bởi

a) | x + $\frac{4}{15}$ | – | – 3,75 | = – | – 2, 15 |
b) | 5x – 4 | = | x + 2 |
c) [| 2x – 1 | + $\frac{1}{2}$ | = $\frac{4}{5}$
d) | x + 4 | + | 9 – x | = 13

0 bình luận về “a) | x + $\frac{4}{15}$ | – | – 3,75 | = – | – 2, 15 | b) | 5x – 4 | = | x + 2 | c) [| 2x – 1 | + $\frac{1}{2}$ | = $\frac{4}{5}$ d) | x + 4 | + | 9”

  2. $a$) `|x + 4/15| – |-3,75| = – |-2,15|`

    `⇔ |x + 4/15| – 3,75 = -2,15`

    `⇔ |x+4/15| = 8/5`

    `⇒` \(\left[ \begin{array}{l}x=\dfrac{4}{3}\\x=\dfrac{-28}{15}\end{array} \right.\) 

       Vậy $x$ $∈$ `{4/3;-28/15}`

    $b$) `|5x-4 | = |x+2|`

    `⇒` \(\left[ \begin{array}{l}5x-4=x+2\\5x-4=-x-2\end{array} \right.\) 

    $⇒$ \(\left[ \begin{array}{l}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{array} \right.\) 

       Vậy $x$ $∈$ `{3/2;1/3}`

    $c$) `||2x-1| + 1/2| = 4/5`

    `⇒` \(\left[ \begin{array}{l}|2x-1| = \dfrac{3}{10}\\|2x-1| = \dfrac{-13}{10}(KTM)\end{array} \right.\) 

    $⇒$ \(\left[ \begin{array}{l}x = \dfrac{13}{20}\\x = \dfrac{7}{20}\end{array} \right.\) 

       Vậy $x$ $∈$ `{13/20;7/20}`

    $d$) `|x+4| + |9-x| = 13`

    Xét $x < -4$

    $⇒ -x – 4 + 9-x = 13$

    $⇔ -2x = 8$

    $⇔ x = -4$

    Xét $-4 ≤ x ≤ 9$

    $⇒ x + 4 + 9-x = 13$

    $⇔ 0x = 0$ 

    Xét $x > 9$

    $⇔ x+4 + x – 9 = 13$

    $⇔ 2x = 18$

    $⇔ x = 9$

       Vậy $-4 ≤ x ≤ 9$

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