\(d,\Leftrightarrow\left|x-2\right|=\left(-2+3\right)\cdot\dfrac{1}{2}=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}+2=\dfrac{5}{2}\\x=-\dfrac{1}{2}+2=\dfrac{3}{2}\end{matrix}\right.\\ e,\Leftrightarrow\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{3}{4}}=\dfrac{z-1}{\dfrac{5}{8}}=\dfrac{x+y-z+1}{\dfrac{3}{2}+\dfrac{3}{4}-\dfrac{5}{8}}=\dfrac{20}{\dfrac{13}{8}}=\dfrac{160}{13}\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{240}{13}\\y=\dfrac{120}{13}\\z=\dfrac{113}{13}\end{matrix}\right.\)
\(f,\Leftrightarrow\left(x-\dfrac{5}{3}\right)\cdot\dfrac{2}{3}=2+\dfrac{1}{8}=\dfrac{17}{8}\\ \Leftrightarrow x-\dfrac{5}{3}=\dfrac{51}{16}\Leftrightarrow x=\dfrac{233}{48}\)
