[tex]\it 8)\ \ 2,5x=\dfrac{x}{2}+4|_{\cdot2} \Rightarrow 5x=x+8 \Rightarrow 5x-x=8 \Rightarrow4x=8\Rightarrow x=2 \\ \\ \\ 16)\ |2x-5|=1 \Rightarrow 2x-5=\pm1 \Rightarrow 2x-5\in\{-1,\ 1\}|_{+5}\Rightarrow \\ \\ \Rightarrow2x\in \{4,\ 6\}|_{:2} \Rightarrow x\in\{2,\ 3\}[/tex]
