Ta có : \(\frac{1}{3}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-.....-\frac{1}{72}-\frac{1}{90}-\frac{1}{100}=x-\frac{5}{13}\)
<=> \(\frac{1}{3}-\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+....+\frac{1}{72}+\frac{1}{90}\right)-\frac{1}{100}=x-\frac{5}{13}\)
<=> \(\frac{1}{3}-\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+.....+\frac{1}{9.10}\right)-\frac{1}{100}=x-\frac{5}{13}\)
<=> \(\frac{1}{3}-\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....+\frac{1}{9}-\frac{1}{10}\right)-\frac{1}{100}=x-\frac{5}{13}\)
<=> \(\frac{1}{3}-\left(\frac{1}{3}-\frac{1}{10}\right)-\frac{1}{100}=x-\frac{5}{13}\)
<=> \(\frac{1}{3}-\frac{1}{3}+\frac{1}{10}-\frac{1}{100}=x-\frac{5}{13}\)
<=> \(\frac{9}{100}=x-\frac{5}{13}\)
<=> \(x=\frac{5}{13}+\frac{9}{100}=\frac{617}{1300}\)