Gọi \(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+....+\frac{1}{17.18}\)

\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{17}-\frac{1}{18}\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{17}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{18}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{18}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{18}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{18}-1-\frac{1}{2}-\frac{1}{3}-.....-\frac{1}{9}\)

\(=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+....+\frac{1}{18}\)

Ta thấy : \(\frac{1}{10}>\frac{1}{19};\frac{1}{11}>\frac{1}{19};\frac{1}{12}>\frac{1}{19};....;\frac{1}{18}>\frac{1}{19}\)

\(\Rightarrow A=\frac{1}{10}+\frac{1}{11}+...+\frac{1}{18}>\frac{1}{19}+\frac{1}{19}+...+\frac{1}{19}\)(có 9 số \(\frac{1}{19}\) )

\(\Rightarrow A>9.\frac{1}{19}=\frac{9}{19}\)(đpcm)

Ta có:  1/1.2 = 1- 1/2 

           1/3.4 = 1/3 - 1/4 

          ...............

             1/19.20 = 1/19 - 1/20 

Cộng vế với vế ta đc: 

    A = 1- 1/20 = 19/20   

\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{17.18}+\frac{1}{19.20}\)

\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{17}-\frac{1}{18}+\frac{1}{19}-\frac{1}{20}\)

\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{17}+\frac{1}{19}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+..+\frac{1}{18}+\frac{1}{20}\right)\)

\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+\frac{1}{20}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{18}+\frac{1}{20}\right)\)

\(A=\left(1+\frac{1}{2}+...+\frac{1}{20}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{9}+\frac{1}{10}\right)\)

\(A=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{19}+\frac{1}{20}\)

\(\frac{A}{B}=1\)

\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{19.20}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}\)

\(\Rightarrow A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{19}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{20}\right)\)

\(\Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{20}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{20}\right)\)

\(\Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{20}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)\)

\(\Rightarrow A=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{20}=B\)

\(\Rightarrow\frac{A}{B}=1\)

Ai giúp đi, làm ơnnnnnnnnnnnnnnnnnnn

\(B=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(B=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(B=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)

\(B=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)

\(B< \frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\)

\(B< \frac{50}{60}\Leftrightarrow B< \frac{5}{6}\)

Ta có : 

\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)

\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)

=> đpcm

Ủng hộ mk nha !!! ^_^

\(\text{Ta có :}\)\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)

\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)

 \(\text{Ta có :}\) \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)

\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)

Ta có :

    \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}-2.\frac{1}{2}-2.\frac{1}{4}-2.\frac{1}{6}-...-2.\frac{1}{50}\)

\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{25}\)

\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\)

    Vậy ......

                         ~Hok tốt~

CỨ TÍNH VẾ TRÁI R ĐC 

1-1/2+1/3-1/4+1/5-1/6+...+1/49-1/50

(1+1/3+1/5+...+1/49)+(1+1/2+1/4+...+1/50)-(1/2+1/4+1/6+...+1/50)-(1/2+1/4+1/6+...+1/50)

(1+1/2+1/3+1/4+...+1/50)-2(1/2+1/4+1/6+..+1/50)

(1+1/2+1/4+1/6+...+1/50)-(1+1/2+1/3+..+1/25)

1/26+1/27+...+1/50

\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)

Ta có: Vế trái:\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+..+\frac{1}{48}+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+..+\frac{1}{24}+\frac{1}{25}\right)\)

\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{25}\)

\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}=\)vế phải

\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{49}+\frac{1}{50}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{49}+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{25}\right)\)

\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+\frac{1}{29}+...+\frac{1}{50}\left(đpcm\right)\)