Tìm số nguyên x sao cho a) 2x chia hết cho x+5. b) x²+2 chia hết cho x+1

$a$) $2x \vdots x+5$

$⇔$ $2x – 2(x+5) \vdots x+5$

$⇔$ $2x – 2x – 10 \vdots x+5$

$⇔$ $-10 \vdots x+5$

$⇒$ $x+5$ ∈ Ư($10$)={$±1;±2;±5;±10$}

$⇔$ $x$ ∈ {$-15;-10;-7;-6;-4;-3;0;5$}

 Vậy $x$ ∈ {$-15;-10;-7;-6;-4;-3;0;5$}

$b$) $x^2 + 2 \vdots x+1$

$⇔$ $x^2 + 2 – x(x+1) \vdots x+1$

$⇔$ $x^2 + 2 – x^2 – x  \vdots x+1$

$⇔$ $ 2-x \vdots x+1$

$⇔$ $2-x + x+1  \vdots x+1$

$⇔$ $3 \vdots x+1$

$⇒$ $x+1$ ∈ Ư($3$)={$±1;±3$}

$⇔$ $x$ ∈ {$-4;-2;0;2$}

 Vậy $x$ ∈ {$-4;-2;0;2$}