Tìm x biết a) |x+1/3|-5=7 b) 3(x-2)+2/5=4

Tìm x biết
a) |x+1/3|-5=7
b) 3(x-2)+2/5=4

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  1. Giải:

    `a, |x + 1/3| – 5 = 7`

    `-> |x + 1/3| = 7 + 5`

    `-> |x + 1/3| = 12`

    `->` \(\left[ \begin{array}{l}x+\dfrac{1}{3} = 12\\x+\dfrac{1}{3} = -12\end{array} \right.\) 

    `->` \(\left[ \begin{array}{l}x = 12 – \dfrac{1}{3} = \dfrac{36}{3} – \dfrac{1}{3} = \dfrac{35}{3}\\x = – 12 – \dfrac{1}{3} = \dfrac{-36}{3} – \dfrac{1}{3} = \dfrac{-37}{3}\end{array} \right.\) 

    Vậy `x ∈ { 35/3 ; -37/3 }`.

    `b, 3(x – 2) + 2/5 = 4`

    `-> 3(x – 2) = 4 – 2/5`

    `-> 3(x – 2) = 20/5 – 2/5`

    `-> 3(x – 2) = 18/5`

    `-> x – 2 = 18/5 : 3`

    `-> x – 2 = 18/5 . 1/3`

    `-> x – 2 = 6/5`

    `-> x = 6/5 + 2`

    `-> x = 6/5 + 10/5`

    `-> x = 16/5`

    Vậy `x = 16/5`.

     

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  2. a) `|x+1/3|-5=7`

    `⇒|x+1/3|=12`

    $⇒\left[ \begin{array}{l}x+\frac{1}{3}=12\\x+\frac{1}{3}=-12\end{array} \right.$

    $⇒\left[ \begin{array}{l}x=\frac{35}{3}\\x=-\frac{37}{3}\end{array} \right.$

    Vậy `x=35/3` hoặc `x=-37/3`

    b) `3(x-2)+2/5=4`

    `⇔3(x-2)=18/5`

    `⇔x-2=6/5`

    `⇔x=16/5`

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