\({\left( {\dfrac{3}{7}} \right)^{2x + 1}} \ge \dfrac{2}{7}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\left( {\dfrac{3}{7}} \right)^{2x + 1}} \ge {\left( {\dfrac{3}{7}} \right)^{{{\log }_{\dfrac{3}{7}}}\dfrac{2}{7}}}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,2x + 1 \le {\log _{\dfrac{3}{7}}}\dfrac{2}{7}\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,2x \le {\log _{\dfrac{7}{3}}}\dfrac{7}{2}-{\log _{\dfrac{7}{3}}}\dfrac{7}{3}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,2x \le {\log _{\dfrac{7}{3}}}\left( {\dfrac{7}{2} \cdot \dfrac{3}{7}} \right)\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x \le \dfrac{1}{2}{\log _{\dfrac{7}{3}}}\dfrac{3}{2}\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,x \le {\log _{\dfrac{7}{3}}}\sqrt {\dfrac{3}{2}} \,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x\, \in \,\left( {-\infty ;{{\log }_{\dfrac{7}{3}}}\sqrt {\dfrac{3}{2}} } \right].\)
Так как \(0 = {\log _{\dfrac{7}{3}}}1 < {\log _{\dfrac{7}{3}}}\sqrt {\dfrac{3}{2}} < {\log _{\dfrac{7}{3}}}\dfrac{7}{3} = 1\), то наибольшее целое решение \(x = 0.\)
Ответ: 0.