Đặt \(A=\frac{\frac{2000}{11}+\frac{2000}{12}+...+\frac{2000}{100}}{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{98}{2}+\frac{99}{1}}\)

\(\Rightarrow A=\frac{2000.\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}\right)}{\left(1+\frac{1}{99}\right)+\left(1+\frac{2}{98}\right)+...+\left(1+\frac{98}{2}\right)+1}\)

\(\Rightarrow A=\frac{2000.\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}\right)}{\frac{100}{99}+\frac{100}{98}+...+\frac{100}{2}+\frac{100}{100}}\)

\(\Rightarrow A=\frac{2000.\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}\right)}{100.\left(\frac{1}{99}+\frac{1}{98}+...+\frac{1}{2}+\frac{1}{100}\right)}\)

\(\Rightarrow A=\frac{20.\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}\right)}{\frac{1}{99}+\frac{1}{98}+...+\frac{1}{2}+\frac{1}{100}}\)

\(\Rightarrow A=\frac{\frac{20}{11}+\frac{20}{12}+..+\frac{20}{100}}{\frac{1}{99}+\frac{1}{98}+..+\frac{1}{2}+\frac{1}{100}}\)

\(\left[\frac{2000}{2000.2006}+\frac{2000}{2006.2012}+...+\frac{2000}{2492.2498}\right]\times\left[\frac{3^2}{8.11}+\frac{3^2}{11.14}+\frac{3^2}{14.17}+...+\frac{3^2}{197.200}\right]\)

\(=\left[\frac{2000}{6}\cdot\left(\frac{1}{2000}-\frac{1}{2006}+...+\frac{1}{2492}-\frac{1}{2498}\right)\right]\times\left[\frac{9}{8.11}+\frac{9}{11.14}+...+\frac{9}{197.200}\right]\)

\(=\left[\frac{2000}{6}\cdot\left(\frac{1}{2000}-\frac{1}{2498}\right)\right]\times\left[\frac{9}{3}\cdot\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+..+\frac{1}{197}-\frac{1}{200}\right)\right]\)

\(=\left[\frac{2000}{6}\cdot\frac{498}{4996000}\right]\times\left[\frac{9}{3}\cdot\left(\frac{1}{8}-\frac{1}{200}\right)\right]\)

\(=\frac{83}{2498}\times\left[\frac{9}{3}\cdot\frac{3}{25}\right]\)

\(=\frac{83}{2498}\times\frac{9}{25}=\frac{747}{62450}\)

\(\frac{A}{B}=\frac{\frac{2000}{1}+\frac{1999}{2}+...+\frac{1}{2000}+2000}{1+\frac{1999}{2}+\frac{1998}{3}+...+\frac{1}{2000}}\)

\(=\frac{\left[\frac{2001}{1}+1\right]+\left[\frac{2001}{2}+1\right]+...+\left[\frac{2001}{2000}+1\right]+2001}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2000}}\)

\(=\frac{2001\left[1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2000}\right]}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2000}}=2001\)

$\ge $ 

là sao?

cho mình hỏi và đề đâu

\(S=2000.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\right)=2000.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\right)\)

  \(=2000.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\right)=2000.\left(1-\frac{1}{100}\right)=20.99=1980\)

\(\left(\frac{x-10}{1994}-1\right)\)+\(\left(\frac{x-8}{1996}-1\right)\)+\(\left(\frac{x-6}{1998}-1\right)\)+\(\left(\frac{x-4}{2000}-1\right)\)+\(\left(\frac{x-2}{2002}-1\right)\)=\(\left(\frac{x-2002}{2}-1\right)\)+\(\left(\frac{x-2000}{4}-1\right)\)+\(\left(\frac{x-1998}{6}-1\right)\)+\(\left(\frac{x-1996}{8}-1\right)\)+\(\left(\frac{x-1994}{10}-1\right)\)

suy ra \(\frac{x-2004}{1994}\)+\(\frac{x-2004}{1996}\)+\(\frac{x-2004}{1998}\)+\(\frac{x-2004}{2000}\)+\(\frac{x-2004}{2002}\)=\(\frac{x-2004}{2}\)+\(\frac{x-2004}{4}\)+\(\frac{x-2004}{6}\)+\(\frac{x-2004}{8}\)+\(\frac{x-2004}{10}\)

suy ra  \(\frac{x-2004}{1994}\)+\(\frac{x-2004}{1996}\)+\(\frac{x-2004}{1998}\)+\(\frac{x-2004}{2000}\)+\(\frac{x-2004}{2002}\)- \(\frac{x-2004}{2}\)- \(\frac{x-2004}{4}\)- \(\frac{x-2004}{6}\)- \(\frac{x-2004}{8}\)- \(\frac{x-2004}{10}\)=0

suy ra (x-2004) . ( \(\frac{1}{1994}\)+\(\frac{1}{1996}\)+\(\frac{1}{1998}\)+\(\frac{1}{2000}\)+\(\frac{1}{2002}\)-\(\frac{1}{2}\)-\(\frac{1}{4}\)-\(\frac{1}{6}\)- \(\frac{1}{8}\)- \(\frac{1}{10}\))=0

Vì  \(\frac{1}{1994}\)+\(\frac{1}{1996}\)+\(\frac{1}{1998}\)+\(\frac{1}{2000}\)+\(\frac{1}{2002}\)-\(\frac{1}{2}\)-\(\frac{1}{4}\)-\(\frac{1}{6}\)- \(\frac{1}{8}\)- \(\frac{1}{10}\) khác 0

nên x-2004=0 suy ra x=2004

em cảm ơn

Giả sử trong 2000 số nguyên dương đã cho không có 2 số nào bằng nhau

\(a_1>a_2>a_3>...>a_{2000}\ge1\)

Khi đó ta có :

\(\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+...+\frac{1}{a_{2000}}\le1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2000}=8,1783...< 12\)

( Mâu thuẫn giả thiết )

Vậy trong 2000 số nguyên dương đã cho có ít nhất 2 số bằng nhau.

Ta có :

\(B=\frac{1999+2000}{2000+2001}=\frac{1999}{2000+2001}+\frac{2000}{2000+2001}\)

VẬY \(\frac{1999}{2000}>\frac{1999}{2000+2001}\)

        \(\frac{2000}{2001}>\frac{2000}{2000+2001}\)

\(\Rightarrow\frac{1999}{2000}+\frac{2000}{2001}>\frac{1999+2000}{2000+2001}\)

\(\Rightarrow A>B\)

CHÚC BN HỌC TỐT #

\(B=\frac{1999+2000}{2000+2001}=\frac{1999}{2000+2001}+\frac{2000}{2000+2001}\)

Ta có: \(\frac{1999}{2000}>\frac{1999}{2000+2001}\)

             \(\frac{2000}{2001}>\frac{2000}{2000+2001}\)

\(\Rightarrow A>B\)

\(A=\frac{1999}{2000}+\frac{2000}{2001}v\text{à}B=\frac{1999+2000}{2000+2001}=\frac{1999}{2000+2001}+\frac{2000}{2000+2001}\)

\(V\text{ì}\frac{1999}{2000}>\frac{1999}{2000+2001};\frac{2000}{2001}>\frac{2000}{2000+2001}\)

Từ đó suy ra A>B