Tìm x: a, $\frac{x-7}{2020}$ + $\frac{x-7}{2021}$ = $\frac{x-7}{2020}$ b, $\frac{1}{2}$x – $\frac{1}{6}$x + 1 = 0

Bởi

Tìm x:
a, $\frac{x-7}{2020}$ + $\frac{x-7}{2021}$ = $\frac{x-7}{2020}$
b, $\frac{1}{2}$x – $\frac{1}{6}$x + 1 = 0

0 bình luận về “Tìm x: a, $\frac{x-7}{2020}$ + $\frac{x-7}{2021}$ = $\frac{x-7}{2020}$ b, $\frac{1}{2}$x – $\frac{1}{6}$x + 1 = 0”

  1. Đáp án:

    `a)`

    ` (x-7)/2020 + (x-7)/2021 = (x-7)/2020`

    ` => (x-7)/2020 + (x-7)/2021 –  (x-7)/2020=0`

    ` => (x-7)(1/2020 + 1/2021 – 1/2020) = 0`

    ` => (x-7). 1/2021 = 0`

    ` => x – 7 = 0`

    ` => x= 7`

    Vậy `  x = 7`

    ` b)`

    ` 1/(2)x – 1/(5)x +1  = 0`

    ` => x (1/2 – 1/5) +1 = 0`

    ` => 3/(10)x + 1 = 0`

    ` => 3/(10)x = -1`

    ` => x= -1 : 3/(10)`

    ` => x = -10/3` 

     

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