a, Ta có : S = \(\frac{1}{50}+\frac{1}{51}+\frac{1}{52}+...+\frac{1}{99}\)
⇔ S = \(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}\right)\)
⇔ \(S=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{98}\right)\)
⇔\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}\) ( 99 số hạng)
⇔ S = \(\left(1-\frac{1}{2}+\frac{1}{3}\right)-\left(\frac{1}{4}-\frac{1}{5}\right)-\left(\frac{1}{6}-\frac{1}{7}\right)-...-\left(\frac{1}{98}-\frac{1}{99}\right)\)
⇔ S = \(\frac{5}{6}-\left(\frac{1}{4}-\frac{1}{5}\right)-\left(\frac{1}{6}-\frac{1}{7}\right)-...-\left(\frac{1}{98}-\frac{1}{99}\right)\)
Mà ta có \(\left(\frac{1}{4}-\frac{1}{5}\right)-\left(\frac{1}{6}-\frac{1}{7}\right)-...-\left(\frac{1}{98}-\frac{1}{99}\right)\) < 0
⇔ \(-\)\(\left(\frac{1}{4}-\frac{1}{5}\right)-\left(\frac{1}{6}-\frac{1}{7}\right)-...-\left(\frac{1}{98}-\frac{1}{99}\right)\) > 0
Như vậy ta được S > \(\frac{5}{6}\) đpcm
b, \(A=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+..+\frac{1}{99}+\frac{1}{100}\) ( 91 số hạng)
Ta có \(\frac{1}{11}>\frac{1}{100};\frac{1}{12}>\frac{1}{100};..;\frac{1}{99}>\frac{1}{100}\)
⇒ \(A>\frac{1}{10}+\frac{1}{100}+\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}+\frac{1}{100}\) (90 số hạng 100)
⇒ A \(>\frac{10}{100}+90.\frac{1}{100}\)
⇒ A > \(\frac{10}{100}+\frac{90}{100}\)
⇒ A > \(\frac{100}{100}=1\)
Vậy ...
