$\begin{array}{l}
y = \dfrac{{\ln x}}{x} \to y' = \dfrac{{\dfrac{1}{x}.x - \ln x}}{{{x^2}}} = \dfrac{{1 - \ln x}}{{{x^2}}}\\
y' = 0 \Leftrightarrow 1 - \ln x = 0 \Leftrightarrow x = e \notin \left[ {7;8} \right]\\
y\left( 7 \right) = \dfrac{{\ln 7}}{7},y\left( 8 \right) = \dfrac{{\ln 8}}{8}\\
 \Rightarrow \mathop {\min }\limits_{\left[ {7;8} \right]} y = y\left( 8 \right) = \dfrac{{\ln 8}}{8}
\end{array}$