Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{1}{1.2}+\frac{1}{3.4}+......+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{3}-....+\frac{1}{49}-\frac{1}{50}=\left(1+\frac{1}{3}+....+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{50}\right)=\left(1+\frac{1}{2}+.....+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)=\left(1+\frac{1}{2}+....+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{25}\right)=\frac{1}{26}+\frac{1}{27}+....+\frac{1}{50}\left(đpcm\right)\)
\(theocaua\Rightarrow A=\frac{1}{26}+\frac{1}{27}+......+\frac{1}{50}>\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\left(5sohang\right)+\frac{1}{40}+\frac{1}{40}+....+\frac{1}{40}\left(10sohang\right)+\frac{1}{50}+\frac{1}{50}+....+\frac{1}{50}\left(10sohang\right)=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=\frac{37}{60}>\frac{35}{60}=\frac{7}{12}\left(1\right)\)
\(A=\frac{1}{26}+\frac{1}{27}+....+\frac{1}{50}< \frac{1}{25}+\frac{1}{25}+...+\frac{1}{25}\left(5sohang\right)+\frac{1}{30}+\frac{1}{30}+....+\frac{1}{30}\left(10sohang\right)+\frac{1}{40}+\frac{1}{40}+.....+\frac{1}{40}\left(10sohang\right)=\frac{1}{4}+\frac{1}{3}+\frac{1}{5}=\frac{47}{60}< \frac{5}{6}=\frac{50}{60}\left(2\right)\) \(\left(1\right);\left(2\right)\Rightarrow\frac{7}{12}< A< \frac{5}{6}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+............+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A=1-\frac{1}{100}\)
\(\Rightarrow A=\frac{99}{100}\)
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 
)
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.
+) \(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(A=\frac{1}{2}+\frac{1}{12}+\frac{1}{30}+...+\frac{1}{9900}\)
\(A=\left(\frac{1}{2}+\frac{1}{12}\right)+\left(\frac{1}{30}+...+\frac{1}{9900}\right)>\frac{1}{2}+\frac{1}{12}.\)
\(\Rightarrow A>\frac{1}{2}+\frac{1}{12}\)
\(\Rightarrow A>\frac{7}{12}\left(1\right).\)
+) \(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=\left(1-\frac{1}{2}+\frac{1}{3}\right)-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{5}{6}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}< \frac{5}{6}\)
\(\Rightarrow A< \frac{5}{6}\left(2\right).\)
Từ \(\left(1\right)và\left(2\right)\Rightarrow\frac{7}{12}< A< \frac{5}{6}\left(đpcm\right).\)
Chúc bạn học tốt!
A = 1 / (1.2) + 1 / (3.4) + ... + 1 / (99.100) > 1 / (1.2) + 1 / (3.4) = 1 / 2 + 1 / 12 = 7 / 12 (1)
A = 1 / (1.2) + 1 / (3.4) + ... + 1 / (99.100) = (1 - 1 / 2) + (1 / 3 - 1 / 4) + ... + (1 / 99 - 100) = (1 - 1 / 2 + 1 / 3) - (1 / 4 - 1 / 5) - (1 / 6 - 1 / 7) - ... - (1 / 98 - 1 / 99) - 1 / 100 < 1 - 1 / 2 + 1 / 3 = 5 / 6 (2)
(1), (2) => 7 / 12 < A < 5 / 6
uikuhikjhkhjjkhjkh