a 1phần4 -(x+1phần2)=3phần5 bà|x-1,4|=2,6

a 1phần4 -(x+1phần2)=3phần5
bà|x-1,4|=2,6

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  1. $\text{a) $\dfrac{1}{4}$ – (x + $\dfrac{1}{2}$) = $\dfrac{3}{5}$}$

    $\text{x + $\dfrac{1}{2}$ = $\dfrac{1}{4}$ – $\dfrac{3}{5}$}$

    $\text{x + $\dfrac{1}{2}$ = $\dfrac{1}{4}$ – $\dfrac{3}{5}$}$

    $\text{x + $\dfrac{1}{2}$ = $\dfrac{5}{20}$ – $\dfrac{12}{20}$}$

    $\text{x + $\dfrac{1}{2}$ = $\dfrac{-7}{20}$ }$

    $\text{x = $\dfrac{-7}{20}$ – $\dfrac{1}{2}$ }$

    $\text{x = $\dfrac{-7}{20}$ – $\dfrac{10}{20}$ }$

    $\text{x = $\dfrac{-17}{20}$ }$

    $\text{b) |x – 1,4| = 2,6}$

    $\text{x – 1,4 = 2,6 hoặc x – 1,4 = -2,6}$

    $\text{TH1: x – 1,4 = 2,6}$

    $\text{        x = 2,6 + 1,4}$

    $\text{        x = 4}$

    $\text{TH2: x – 1,4 = -2,6}$

    $\text{        x = -2,6 + 1,4}$

    $\text{        x = -1,2}$

    $\text{Vậy x ∈ {4; -1,2}}$

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  2. `a)1/4-(x+1/2)=3/5`

    `→x+1/2=1/4-3/5`

    `→x+1/2=-7/20`

    `→x=-7/20-1/2`

    `→x=-17/20`

    Vậy `x=-17/20`

    `b)|x-1,4|=2,6`

    `→` \(\left[ \begin{array}{l}x-1,4=2,6\\x-1,4=-2,6\end{array} \right.\) 

    `→` \(\left[ \begin{array}{l}x=4\\x=-1,2\end{array} \right.\) 

    Vậy `x∈{4;-1,2}`

     

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