Tìm x: a. (x-3).(2y+1)=5 b. (x-2).(1-3y)=7 Giúp mik với nha mn!!!

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

   \begin{array}{|c|c|c|c|c|c|c|c|c|c|}\hline x-3&1&5&-1&-5\\\hline 2y+1&5&1&-5&-1\\\hline x&4&8&2&-2\\\hline y&2&0&-3&-1\\\hline\end{array}

   `b. (x-2)(1-3y)=7 `

   \begin{array}{|c|c|c|c|c|c|c|c|c|c|}\hline x-2&1&7&-1&-7\\\hline 1-3y&7&1&-7&-1\\\hline x&3&9&1&-5\\\hline y&-2&0&\cfrac83&\cfrac13\\\hline\end{array}

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2. Đáp án:

   `↓↓↓`

   Giải thích các bước giải:

   `a) (x – 3) (2y + 1( = 5`

   `-> x – 3, 2y + 1 ∈ Ư (5) = {±1; ±5}`

   * `x – 3 ∈ Ư (5) = {±1; ±5} (x ∈ Z)`

   `-> x – 3 = 1 -> x = 4`

   `-> x – 3 = -1 -> x = 2`

   `-> x – 3 = 5 -> x = 8`

   `-> x – 3 = -5 -> x = -2`

   * `2y + 1 ∈ Ư (5) = {±1; ±5} (y ∈ Z)`

   `-> 2y + 1 = 1 -> y = 0`

   `-> 2y + 1 = -1 -> y = -1`

   `-> 2y – 1 = 5 -> y = 3`

   `-> 2y  – 1 = -5 -> y = -2`

   Vậy ….

   `b) (x – 2) (1 – 3y) = 7`

   `-> x – 2, 1 – 3y ∈ Ư (7) = {±1; ±7}`

   * `x – 2 ∈ Ư (7) = {±1; ±7} (x ∈ Z)`

   `-> x – 2 = 1 -> x = 3`

   `-> x – 2 = -1 -> x = 1`

   `-> x – 2 = 7 -> x = 9`

   `-> x – 2 = -7 -> x = -5`

   * `1 – 3y ∈ Ư (7) = {±1; ±7} (y ∈ Z)`

   `-> 1 – 3y = 1 -> y = 0`

   `-> 1 – 3y = -1 -> y = 2/3`

   `-> 1 – 3y = 7 -> y = -2`

   `-> 1 – 3y = -7 -> y = 8/3`

   Vậy …

    

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