\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+....+\dfrac{1}{24\times25}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{24}-\dfrac{1}{25}\)

\(=1-\dfrac{1}{25}\)

\(=\dfrac{24}{25}\)

Nhanh giúp mình với cả nhà ơi

#)Giải :

\(C=\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{15.20}\right)\div\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}\right)\)

\(C=\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}\right)\div\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}\right)\)

\(C=\left(1-\frac{1}{20}\right)\div\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}\right)\)

\(C=\frac{19}{20}\div\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}\right)\)

Còn vế kia thì chịu @@

\(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{100}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{100}\right)\)

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

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

\(=\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{75}\right)+\left(\dfrac{1}{76}+\dfrac{1}{77}+...+\dfrac{1}{100}\right)\)

Ta có:

\(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{75}>\dfrac{1}{75}+\dfrac{1}{75}+...+\dfrac{1}{75}=\dfrac{25}{75}=\dfrac{1}{3}\)

\(\dfrac{1}{76}+\dfrac{1}{77}+...+\dfrac{1}{100}>\dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}=\dfrac{25}{100}=\dfrac{1}{4}\)

\(\Rightarrow A>\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{7}{12}\) (1)

Lại có:

\(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{75}< \dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}=\dfrac{25}{50}=\dfrac{1}{2}\)

\(\dfrac{1}{76}+\dfrac{1}{77}+...+\dfrac{1}{100}< \dfrac{1}{75}+\dfrac{1}{75}+...+\dfrac{1}{75}=\dfrac{25}{75}=\dfrac{1}{3}\)

\(\Rightarrow A< \dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\) (2)

Từ (1) và (2) suy ra \(\dfrac{7}{12}< A< \dfrac{5}{6}\)

\(A=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{99}-\frac{1}{100}\)

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

\(=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)

Do \(\frac{1}{51}>\frac{1}{52}>...>\frac{1}{100}\Rightarrow A=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}>25\cdot\frac{1}{80}+25\cdot\frac{1}{100}=\frac{7}{12}\)

và \(A

olm lag kinh đang làm lag thoát ra mất tiêu

-------đề đúng------------

Ta có : 

Vậy ĐPCM

a) 1/1x5 + ... + 1/21x25

= 4 x (1-1/5 + 1/5 - 1/9 + ... + 1/21 - 1/25)

= 1/4 x (1 - 1/25)

= 1/4 x 24/25

= 6/25

ai làm nhanh mik cho

mik cho

haizzz đáng tiếc tôi muốn ns là:  ko bao f và đừng mong chờ OK