$|x+9|.2=10$
$⇒ |x+9| = 5$
$⇒$ \(\left[ \begin{array}{l}x+9=5\\x+9=-5\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}x=-4\\x=-14\end{array} \right.\)
Vậy $x$ $∈$ `{-4;-14}`
Ta có:
$\left \{ {{x \vdots 12} \atop {x \vdots 10}} \right.$
$⇒$ $x$ $∈$ `BC(10;12)`
$⇒$ $x$ là bội của `BCNNNN(10;12)`
$10=2.5$
$12 = 2^2.3$
$⇒$ `BCNNNN(10;12)=2^2.3.5=60$
$⇒$ $x$ là `B(60)={0;±60;±120;±180;±240;..........}`
Mà $-200 ≤ x ≤ 200$
$⇒$ $x$ $∈$ `{0;±60;±120;±180}`
Vậy $x$ $∈$ `{0;±60;±120;±180}`
