$a$.$35-5$($x+3$)=|$-15$|
⇔$35-5$($x+3$)=$15$
⇔ $5$($x+3$)=$35-15$
⇔ $5$($x+3$)=$20$
⇔ $x+3$ = $20:5$
⇔ $x+3$ = $4$
⇔ $x =1$
Vậy $x=1$
$b$.|$x-7$|-$(-15)^{0}$=|$-6$|
⇔|$x-7$|-$1$=$6$
⇔ |$x-7$| = $7$
⇒\(\left[ \begin{array}{l}x-7=7\\x-7=-7\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=14\\x=0\end{array} \right.\)
Vậy $x$ ∈ {$0;14$}
$c$.$5x+17=x-47$
⇔$17+47 = x - 5x$
⇔$64 = -4x$
⇔ $-x = -16$
⇒ $x=16$
Vậy $x=16$
