tim x biet a, 1/3-|5/4-2x|=1/4 b, 1/2-|x+1/5|=1/3 c, 3/4-|2x+1|=7/8 d, 2|2x-3|=1/2 e, 7,5-3|5-2x|=-4,5

Bởi

tim x biet
a, 1/3-|5/4-2x|=1/4
b, 1/2-|x+1/5|=1/3
c, 3/4-|2x+1|=7/8
d, 2|2x-3|=1/2
e, 7,5-3|5-2x|=-4,5

0 bình luận về “tim x biet a, 1/3-|5/4-2x|=1/4 b, 1/2-|x+1/5|=1/3 c, 3/4-|2x+1|=7/8 d, 2|2x-3|=1/2 e, 7,5-3|5-2x|=-4,5”

  1. `a, 1/3-|5/4-2x|=1/4`

     `⇔ |5/4 – 2x| = 1/{12}`

    `⇒` \(\left[ \begin{array}{l}2x=\dfrac{7}{6}\\2x=\dfrac{4}{3}\end{array} \right.\) 

    $⇒$ \(\left[ \begin{array}{l}x=\dfrac{7}{12}\\x=\dfrac{2}{3}\end{array} \right.\) 

       Vậy $x$ $∈$ `{7/{12};2/3}`

    $b$) `1/2-|x+1/5|=1/3`

    `⇔ |x+1/5| = 1/6`

    `⇒` \(\left[ \begin{array}{l}x=\dfrac{-1}{30}\\x=\dfrac{-11}{30}\end{array} \right.\) 

       Vậy $x$ $∈$ `{-1/{30};{-11}/{30}}`

    `c`) `3/4-|2x+1|=7/8`

    `⇔ |2x+1| = -1/8`

    `⇒` `x` `∈` $∅$ vì $|2x+1|$ $≥$ $0$ $∀$ $x$

      Vậy `x` `∈` $∅$

    $d$) `2|2x-3| = 1/2`

    `⇔ |2x-3| =1/4`

    `⇒` \(\left[ \begin{array}{l}2x=\dfrac{13}{4}\\2x=\dfrac{11}{4}\end{array} \right.\) 

    $⇒$ \(\left[ \begin{array}{l}x=\dfrac{13}{8}\\x=\dfrac{11}{8}\end{array} \right.\) 

       Vậy $x$ $∈$ `{{13}/8;{11}/8}`

    $e$) `7,5 – 3|5-2x| = -4,5`

    $⇔ 3|5-2x| =12$

    $⇔ |5-2x| = 4$

    $⇒$ \(\left[ \begin{array}{l}2x=1\\2x=9\end{array} \right.\) 

    $⇔$ \(\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=\dfrac{9}{2}\end{array} \right.\) 

       Vậy $x$ $∈$ `{1/2;9/2}`

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