\(A=\frac{34}{7.13}+\frac{51}{13.22}+\frac{85}{22.37}+\frac{68}{37.49}\)
\(=\frac{17.2}{7.13}+\frac{17.3}{13.22}+\frac{17.5}{22.37}+\frac{17.4}{37.49}\)
\(=17\left(\frac{2}{7.13}+\frac{3}{13.22}+\frac{5}{22.37}+\frac{4}{37.49}\right)\)
\(=\frac{17}{3}\left(\frac{6}{7.13}+\frac{9}{13.22}+\frac{15}{22.37}+\frac{12}{37.49}\right)\)
\(=\frac{17}{3}\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{13}-\frac{1}{22}+...+\frac{1}{37}-\frac{1}{49}\right)\)
\(=\frac{17}{3}\left(\frac{1}{7}-\frac{1}{49}\right)\)\(=\frac{17}{3}.\frac{6}{49}=\frac{17.2}{49}=\frac{34}{49}\)
Có : \(\frac{17}{24}=\frac{34}{48}\)
Vì 48 < 49 => \(\frac{34}{48}>\frac{34}{49}\). Hay \(\frac{17}{24}>A\)
