\(a)\,\,|x-3|=2\\ \to\left[ \begin{array}{l}x-3=2\\x-3=-2\end{array} \right. \\ \to\left[ \begin{array}{l}x=2+3\\x=-2+3\end{array} \right.\\\to\left[ \begin{array}{l}x=5\\x=1\end{array} \right. \\ \text{- Vậy}\,\,x \in \{5;1\} \\ \, \\ b)\,\,|5-x|=10 \\ \to\left[ \begin{array}{l}5-x=10\\5-x=-10\end{array} \right.\\\to\left[ \begin{array}{l}5-10=x\\5+10=x\end{array} \right.\\\to\left[ \begin{array}{l}-5=x\\15=x\end{array} \right. \\ \text{- Vậy}\,\,x \in \{-5;15\} \\ \, \\ c)\,\,|2x-5|=3 \\ \to\left[ \begin{array}{l}2x-5=3\\2x-5=-3\end{array} \right.\\\to\left[ \begin{array}{l}2x=8\\2x=2\end{array} \right.\\\to\left[ \begin{array}{l}x=4\\x=1\end{array} \right. \\ d)\,\, 7\,\,\vdots\,\,x+2 \\ \to x+2 \in Ư(7)=\{\pm1;\pm7\}\\\to x \in \{-9;-3;-1;5\} \\ e)\,\,5\,\,\vdots\,\, x-3 \\ \to x-3 \in Ư(5)=\{\pm1;\pm5\} \\ \to x \in \{-2;2;5;8\} \\ g)\,\,13\,\,\vdots\,\,2x+1 \\ \to 2x+1 \in Ư(13)=\{\pm1;\pm13\}\\\to2x \in \{-14;-2;0;12\}\\\to x \in \{-7;-1;0;6\}\)
