(3$\frac {1}{2}$ – 2x) : 3 $\frac{1}{3}$ = 7$\frac{1}{3}$

(3$\frac {1}{2}$ – 2x) : 3 $\frac{1}{3}$ = 7$\frac{1}{3}$

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  1. $(3\dfrac{1}{2}-2x):3\dfrac{1}{3}=7\dfrac{1}{3} \\⇔3\dfrac{1}{2}-2x=7\dfrac{1}{3}.3\dfrac{1}{3} \\⇔\dfrac{7}{2}-2x=\dfrac{22}{3}.\dfrac{10}{3} \\⇔\dfrac{7}{2}-2x=\dfrac{220}{9} \\⇔2x=\dfrac{7}{2}-\dfrac{220}{9} \\⇔2x=\dfrac{-377}{18} \\⇔x=\dfrac{-377}{36}$

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  2. Đáp án:

     

    Giải thích các bước giải:

     (3$\dfrac{1}{2} – 2x) : 3\dfrac{1}{3} = 7\dfrac{1}{7}$

    ($\dfrac{7}{2} – 2x) : \dfrac{7}{3} = \dfrac{15}{7}$

    ($\dfrac{7}{2} – 2x) = \dfrac{15}{7} × \dfrac{7}{3}$

    $\dfrac{7}{2} – 2x = 5$

    2x = $\dfrac{7}{2} – 5$

    $2x$ $=$ $\dfrac{-3}{2}$

      $x$ $=$ $\dfrac{-3}{4}$

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