\(\left|1-\dfrac{x}{5}-\dfrac{4}{3}\right|=\dfrac{53}{15}\Leftrightarrow\left|\dfrac{-1}{3}-\dfrac{x}{5}\right|=\dfrac{53}{15}\)

th1: \(\dfrac{-1}{3}-\dfrac{x}{5}\ge0\Leftrightarrow-\dfrac{x}{5}\ge\dfrac{1}{3}\Leftrightarrow x\le\dfrac{1}{3}.-5=\dfrac{-5}{3}\)

\(\Rightarrow\left|\dfrac{-1}{3}-\dfrac{x}{5}\right|=\dfrac{53}{15}\Leftrightarrow\dfrac{-1}{3}-\dfrac{x}{5}=\dfrac{53}{15}\Leftrightarrow-\dfrac{x}{5}=\dfrac{53}{15}+\dfrac{1}{3}\)

\(\Leftrightarrow-\dfrac{x}{5}=\dfrac{58}{15}\Leftrightarrow x=\dfrac{58}{15}.-5=\dfrac{-58}{3}\left(tmđk\right)\)

th2: \(\dfrac{-1}{3}-\dfrac{x}{5}< 0\Leftrightarrow-\dfrac{x}{5}< \dfrac{1}{3}\Leftrightarrow x>\dfrac{1}{3}.-5=\dfrac{-5}{3}\)

\(\Rightarrow\left|\dfrac{-1}{3}-\dfrac{x}{5}\right|=\dfrac{53}{15}\Leftrightarrow-\left(\dfrac{-1}{3}-\dfrac{x}{5}\right)=\dfrac{53}{15}\Leftrightarrow\dfrac{1}{3}+\dfrac{x}{5}=\dfrac{53}{15}\)

\(\Leftrightarrow\dfrac{x}{5}=\dfrac{53}{15}-\dfrac{1}{3}\Leftrightarrow\dfrac{x}{5}=\dfrac{16}{5}\Leftrightarrow x=\dfrac{16}{5}.5=16\left(tmđk\right)\)

vậy \(x=\dfrac{-58}{3};x=16\)