\(\frac{1}{1.2}+\frac{1}{3.4}+......+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{3}-....+\frac{1}{49}-\frac{1}{50}=\left(1+\frac{1}{3}+....+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{50}\right)=\left(1+\frac{1}{2}+.....+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)=\left(1+\frac{1}{2}+....+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{25}\right)=\frac{1}{26}+\frac{1}{27}+....+\frac{1}{50}\left(đpcm\right)\)

\(theocaua\Rightarrow A=\frac{1}{26}+\frac{1}{27}+......+\frac{1}{50}>\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\left(5sohang\right)+\frac{1}{40}+\frac{1}{40}+....+\frac{1}{40}\left(10sohang\right)+\frac{1}{50}+\frac{1}{50}+....+\frac{1}{50}\left(10sohang\right)=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=\frac{37}{60}>\frac{35}{60}=\frac{7}{12}\left(1\right)\)

\(A=\frac{1}{26}+\frac{1}{27}+....+\frac{1}{50}< \frac{1}{25}+\frac{1}{25}+...+\frac{1}{25}\left(5sohang\right)+\frac{1}{30}+\frac{1}{30}+....+\frac{1}{30}\left(10sohang\right)+\frac{1}{40}+\frac{1}{40}+.....+\frac{1}{40}\left(10sohang\right)=\frac{1}{4}+\frac{1}{3}+\frac{1}{5}=\frac{47}{60}< \frac{5}{6}=\frac{50}{60}\left(2\right)\) \(\left(1\right);\left(2\right)\Rightarrow\frac{7}{12}< A< \frac{5}{6}\)

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

\(\Rightarrow A=1-\frac{1}{100}\)

\(\Rightarrow A=\frac{99}{100}\)

Vì \(\frac{7}{12}< \frac{99}{100}< \frac{5}{6}\Rightarrow\frac{7}{12}< A< \frac{5}{6}\)               ĐPCM

( Bài này ko ai lm thì t lm cho )

ủa làm sai hay s ik

Hướng dẫn

Áp dụng BĐT để giải

~ Ủng hộ nhé

A = 1/1.2 + 1/3.4 + 1/5.6 + ... + 1/99.100

A = 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + ... + 1/99 - 1/100

A = ( 1 + 1/3 + 1/5 + ... + 1/99) - ( 1/2 + 1/4 + 1/6 + ... + 1/100)

A = ( 1 + 1/2 + 1/3 + 1/4 + .... + 1/99 + 1/100) - 2.( 1/2 + 1/4 + 1/6 + ....+ 1/100)

A = ( 1 + 1/2 + 1/3 + 1/4 + .... + 1/99 + 1/100) - ( 1 + 1/2 + 1/3 + ... + 1/50)

A = 1/51 + 1/52 + .... + 1/100

A = ( 1/51 + 1/52 + ... + 1/75) + ( 1/76 + 1/77 + ... + 1/100)

A > 1/75 . 25 + 1/100 . 25

A > 1/3 + 1/4 = 7/12 (1)

A < 1/50 . 25 + 1/75 x 25

a < 1/2 + 1/3 = 5/6 (2)

Từ (1) và (2) => 7/12 < A < 5/6

Chứng tỏ 7/12 < A < 5/6

A = 1/1.2 + 1/3.4 + 1/5.6 + ... + 1/99.100

A = 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + ... + 1/99 - 1/100

A = ( 1 + 1/3 + 1/5 + ... + 1/99) - ( 1/2 + 1/4 + 1/6 + ... + 1/100)

A = ( 1 + 1/2 + 1/3 + 1/4 + .... + 1/99 + 1/100) - 2.( 1/2 + 1/4 + 1/6 + ....+ 1/100)

A = ( 1 + 1/2 + 1/3 + 1/4 + .... + 1/99 + 1/100) - ( 1 + 1/2 + 1/3 + ... + 1/50)

A = 1/51 + 1/52 + .... + 1/100

A = ( 1/51 + 1/52 + ... + 1/75) + ( 1/76 + 1/77 + ... + 1/100)

A > 1/75 . 25 + 1/100 . 25

A > 1/3 + 1/4 = 7/12 (1)

A < 1/50 . 25 + 1/75 x 25

a < 1/2 + 1/3 = 5/6 (2)

Từ (1) và (2) => 7/12 < A < 5/6

Chứng tỏ 7/12 < A < 5/6

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

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

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

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

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

\(=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)

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

Ta có: \(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}>\frac{1}{75}+\frac{1}{75}+...+\frac{1}{75}=25\cdot\frac{1}{75}=\frac{25}{75}=\frac{1}{3}\)

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

\(\Rightarrow A>\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\left(1\right)\)

Lại có: \(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}< \frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}=25\cdot\frac{1}{50}=\frac{25}{50}=\frac{1}{2}\)

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

\(\Rightarrow A< \frac{1}{2}+\frac{1}{3}=\frac{5}{6}\left(2\right)\)

Từ (1) và (2) => đpcm

\(\Rightarrow\)(1/1.2) + ( 1/ 3.4) + (1/.6) +...+(1/99.100)

\(\Rightarrow\)(\(\frac{1}{1}\)-1/2 +1/3 -1/4 +...+ 1/99 - 1/100)

\(\Rightarrow\)( 1 - 1/100)

\(=\)99/100

Ta có \(\frac{7}{12}\)=0,5833 

           \(\frac{99}{100}\)=0,99

          \(\frac{5}{6}\)=0,8333

Vì 0,99 > 0,8333 > 0,58333

\(\)\(\Leftrightarrow\)\(\frac{99}{100}\)>\(\frac{5}{6}\)>\(\frac{7}{12}\)

Vậy A lớn nhất trong cả 3 số không phải như điều cần chứng minh.