Question

Tìm số nguyên \(x\) sao cho:

\(3x-4\) chia hết cho \(x+1\)

Answer 1

3x - 4 = 3x + 3 - 7 = 3(x + 1) - 7

Để (3x - 4) ⋮ (x + 1) thì 7 ⋮ (x + 1)

⇒ x + 1 ∈ Ư(7) = {-7; -1; 1; 7}

⇒ x ∈ {-8; -2; 0; 6}

Answer 2

Ta có :3x - 4 ⋮ x+1

=> 3x - 4 - [3(x+1)] ⋮ x+1

=> 3x - 4 - (3x + 3) ⋮ x+1

=> 3x - 4 -3x - 3 ⋮ x+1

=> -7 ⋮ x+1

=> x + 1 = -7;-1;1;7 => x = -8;-2;0;6

Answer 3

3x - 4 = 3x + 3 - 7 = 3(x + 1) - 7

Để (3x - 4) ⋮ (x + 1) thì 7 ⋮ (x + 1)

⇒ x + 1 ∈ Ư(7) = {-7; -1; 1; 7}

⇒ x ∈ {-8; -2; 0; 6}

Answer 4

Ta có $3x+4⋮x-3$

$\Rightarrow 3\left(x-3\right)+13⋮x-3$

$\Rightarrow 13⋮x-3$

$\Rightarrow x-3\in Ư\left(13\right)=\left\{-13;-1;1;13\right\}$

$\Rightarrow x\in \left\{-10;2;4;16\right\}$  ( thỏa mãn x nguyên )

Vậy $x\in \left\{-10;2;4;16\right\}$