\(\dfrac{{x + 5}}{{7x + 11}} = \dfrac{{x + 5}}{{6x + 1}}\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{7x + 11 \ne 0,\,\,\,\,\,6x + 1 \ne 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{\left( {x + 5} \right)\left( {7x + 11} \right) — \left( {x + 5} \right)\left( {6x + 1} \right) = 0}\end{array}} \right.\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x \ne — \dfrac{{11}}{7};\,\,\,\,x \ne — \dfrac{1}{6}}\\{\left( {x + 5} \right)\left( {7x + 11 — 6x — 1} \right) = 0}\end{array}\,\,\,\,\, \Leftrightarrow } \right.\)
\[ \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x \ne — \dfrac{{11}}{7};\,\,\,\,x \ne — \dfrac{1}{6}\,\,}\\{\left( {x + 5} \right)\left( {x + 10} \right) = 0}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x \ne — \dfrac{{11}}{7};\,\,\,\,x \ne — \dfrac{1}{6}}\\{{x_1} = — 5,\,\,\,\,{x_2} = — 10}\end{array}\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{x_1} = — 5,\,\,\,\,{x_2} = — 10.} \right.\]
Наибольший из найденных корней – 5.
Ответ: – 5.