a,
$|3-x|=7$
$⇔\left[\begin{array}{l}3-x=-7\\3-x=7\end{array}\right.⇔\left[\begin{array}{l}x=10\\x=-4\end{array}\right.$
Vậy $x \in \{-4;10\}$
b,
$|x-2|-13=-3.(-5)$
$⇔|x-2|-13=15$
$⇔|x-2|=28$
$⇔\left[\begin{array}{l}x-2=-28\\x-2=28\end{array}\right.⇔\left[\begin{array}{l}x=-26\\x=30\end{array}\right.$
Vậy $x \in \{-26;30\}$
c,
$|x^2-5|=(-2)^2$
$⇔|x^2-5|=4$
$⇔\left[\begin{array}{l}x^2-5=-4\\x^2-5=4\end{array}\right.⇔\left[\begin{array}{l}x^2=1\\x^2=9\end{array}\right.⇔\left[\begin{array}{l}x=±1\\x=±3\end{array}\right.$
Vậy $x \in \{±1;±3\}$
d,
$|x-3|+|2x-6|=15$
$⇔|x-3|+2|x-3|=15$
$⇔3|x-3|=15$
$⇔|x-3|=5$
$⇔\left[\begin{array}{l}x-3=-5\\x-3=5\end{array}\right.⇔\left[\begin{array}{l}x=-2\\x=8\end{array}\right.$
Vậy $x \in \{-2;8\}$
e,
$|x-2|+|2-x|=8$
$⇔|x-2|+|x-2|=8$
$⇔2|x-2|=8$
$⇔|x-2|=4$
$⇔\left[\begin{array}{l}x-2=-4\\x-2=4\end{array}\right.⇔\left[\begin{array}{l}x=-2\\x=6\end{array}\right.$
Vậy $x \in \{-2;6\}$
