Bài 2: Tìm x $a,a\dfrac{1}{4}+x=\dfrac{-2}{3}$ $b,(30\%-\dfrac{2}{5}.x):\dfrac{-1}{2}=1\dfrac{2}{3}$ $c,0,5-\dfrac{2}{3}x=\dfrac{-7}{12}$

Bài 2: Tìm x
$a,a\dfrac{1}{4}+x=\dfrac{-2}{3}$ $b,(30\%-\dfrac{2}{5}.x):\dfrac{-1}{2}=1\dfrac{2}{3}$
$c,0,5-\dfrac{2}{3}x=\dfrac{-7}{12}$ $d,(2x+\dfrac{3}{4}).\dfrac{7}{9}=\dfrac{5}{18}$

0 bình luận về “Bài 2: Tìm x $a,a\dfrac{1}{4}+x=\dfrac{-2}{3}$ $b,(30\%-\dfrac{2}{5}.x):\dfrac{-1}{2}=1\dfrac{2}{3}$ $c,0,5-\dfrac{2}{3}x=\dfrac{-7}{12}$”

  1. a)

    $⇒x=\frac{-2}{3}-\frac{1}{4}$
    $⇒x=\frac{-11}{12}$
    b)

    $⇒(\frac{3}{10}-\frac{2}{5}.x)=\frac{-5}{6}$

    $⇒\frac{2}{5}.x=\frac{17}{15}$

    $⇒x=\frac{6}{17}$

    c)

    $⇒(\frac{1}{2}+\frac{2}{3}x=\frac{-7}{12}$

    $⇒\frac{2}{3}x=\frac{-13}{12}$

    $⇒x=\frac{-13}{8}$

    d)

    $⇒(2x+\frac{3}{4})=\frac{5}{14}$

    $⇒2x=\frac{-11}{28}$
    $⇒x=\frac{-11}{56}$

     

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