Bài $3$
$a$) $125 - 5 \times |x-10| = 100$
$⇔ 5 \times |x-10| = 25$
$⇔ |x-10| = 5$
$⇒$ \(\left[ \begin{array}{l}x-10=5\\x-10=-5\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}x=15\\x=5\end{array} \right.\)
Vậy $x$ $∈$ `{5;15}`
$b$) $|x:8 + 12| \times 5 = 60$
$⇔ |x:8 + 12| = 12$
$⇒$ \(\left[ \begin{array}{l}x:8+12=12\\x:8+12=-12\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}x:8=0\\x:8=-24\end{array} \right.\)
$⇔$ \(\left[ \begin{array}{l}x=0\\x=-192\end{array} \right.\)
Vậy $x$ $∈$ `{-192;0}`
$c$) $1 + 2 + 3 + .... + x=5050$
$⇒ \dfrac{(x+1).[(x-1):1+1]}{2} = 5050$
$⇔ (x+1).x = 10100$
$⇒$ \(\left[ \begin{array}{l}x=100\\x=-101\end{array} \right.\)
Mà $x$ $∈$ $N$ $⇒$ $x=100$
Vậy $x=100$.
