Bài làm

x = \(\frac{20}{21}+\frac{21}{22}+\frac{22}{23}+\frac{23}{20}\)

x = 1 + 1 + 1 + 1 + \((\)\(\frac{3}{20}-\frac{1}{21}-\frac{1}{22}-\frac{1}{23})\)

Ta thấy 0 < \(\frac{3}{20}-\frac{1}{21}-\frac{1}{22}-\frac{1}{23}\)

\(\Rightarrow\) 1 + 1 + 1 + 1 + \((\frac{3}{20}-\frac{1}{21}-\frac{1}{22}-\frac{1}{23})\)> 4

\(\Rightarrow\)x > 4

Đặt  \(B=\frac{1}{20}+\frac{1}{200}+\frac{1}{200}+....+\frac{1}{200}< C=\frac{1}{20}+\frac{1}{21}+\frac{1}{22}+....+\frac{1}{200}\)

Số các phân số \(\frac{1}{200}\)có trong \(B\)là :

 ( 200 - 21 ) :1 + 1 = 180 ( phân số )

Nên \(B=\frac{1}{20}+180.\frac{1}{200}=\frac{1}{20}+\frac{9}{10}>\frac{9}{10}\)

Do đó , \(C>B>\frac{9}{10}\)nên \(C>\frac{9}{10}\)

Vậy \(C>\frac{9}{10}\left(ĐPCM\right)\)

ta có C= 1/20 +1/21+1/22+...+1/200

C > 1/200 + 1/200 +1/200 +...+ 1/200(181 ps)

C > 181/200 > 180/200 = 9/10

=> C > 9/10

Ta có : \(S=\frac{1}{20}+\frac{1}{21}+\frac{1}{22}+...+\frac{1}{199}+\frac{1}{200}\)

\(\Rightarrow S>\frac{1}{200}+\frac{1}{200}+\frac{1}{200}+...+\frac{1}{200}\) ( 181 phân số )

\(\Rightarrow S>\frac{181}{200}>\frac{180}{200}=\frac{9}{10}\)

\(\Rightarrow S>\frac{9}{10}\)       \(\Rightarrowđpcm\)

C = 120120 + 121121 + 122122 + ... + 12001200

⇒ CC> 12001200 + 12001200 + 12001200 + ...... + 12001200 ( 181181 phân số )

⇒ CC > 181200181200 > 180200180200 = 910910

⇒ CC >910

\(B=\left(1+4+4^2\right)+...+4^{36}\left(1+4+4^2\right)\)

\(=21\left(1+...+4^{36}\right)⋮21\)