n, $\frac{22}{15}$.x + $\frac{1}{3}$ = |$\frac{-2}{3}$ + $\frac{1}{5}$ |

n, $\frac{22}{15}$.x + $\frac{1}{3}$ = |$\frac{-2}{3}$ + $\frac{1}{5}$ |

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  1. $\frac{22}{15}$ . x + $\frac{1}{3}$ = |$\frac{-2}{3}$ + $\frac{1}{5}$|

    $\frac{22}{15}$ . x + $\frac{1}{3}$ = |$\frac{-7}{15}$|

    $\frac{22}{15}$ . x + $\frac{1}{3}$ = $\frac{7}{15}$

    $\frac{22}{15}$ . x = $\frac{7}{15}$ – $\frac{1}{3}$ 

    $\frac{22}{15}$ . x = $\frac{2}{15}$

    x = $\frac{2}{15}$ : $\frac{22}{15}$ 

    x = $\frac{1}{11}$ 

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  2. Đáp án:

    $\frac{22}{15}$.x+$\frac{1}{3}$=| $\frac{-2}{3}$+$\frac{1}{5}$|

    ⇒ $\frac{22}{15}$.x+$\frac{1}{3}$=|$\frac{-7}{15}$|

    ⇒ $\frac{22}{15}$.x+$\frac{1}{3}$=$\frac{7}{15}$

    ⇔ $\frac{22}{15}$.x=$\frac{7}{15}$-$\frac{1}{3}$

    ⇔ $\frac{22}{15}$.x=$\frac{2}{15}$

    ⇔ x=$\frac{2}{15}$÷$\frac{22}{15}$

    ⇔ x=$\frac{1}{11}$

    Vậy x=$\frac{1}{11}$

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