Ta có:

 \(c=\)\(\frac{1}{101}\)\(+\)\(\frac{1}{102}\)\(+\)\(\frac{1}{103}\)\(+\)...\(+\)\(\frac{1}{200}\)

\(c=\)(\(\frac{1}{101}\)\(+\)\(\frac{1}{102}\)\(+\)...\(+\)\(\frac{1}{120}\))\(+\)(\(\frac{1}{121}\)\(+\)\(\frac{1}{122}\)\(+\)...\(+\)\(\frac{1}{150}\))\(+\)(\(\frac{1}{151}\)\(+\)\(\frac{1}{152}\)\(+\)...\(+\)\(\frac{1}{180}\))\(+\)(\(\frac{1}{181}\)\(+\)\(\frac{1}{182}\)\(+\)...\(+\)\(\frac{1}{200}\))>20\(.\)\(\frac{1}{120}\)\(+\)30\(.\)\(\frac{1}{150}\)\(+\)30\(.\)\(\frac{1}{180}\)\(+\)20\(.\)\(\frac{1}{200}\)= \(\frac{1}{6}+\frac{1}{5}\)\(+\)\(\frac{2}{6}+\frac{1}{10}\)= \(\frac{19}{30}\)=\(\frac{76}{120}\)> \(\frac{75}{120}\)=\(\frac{5}{8}\)

=>\(c\)>\(\frac{5}{8}\)(đpcm)

_Hok tốt_

Cảm ơn nhiều !

Ta có: \(C=\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{200}\)

\(=\left(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{120}\right)+\left(\dfrac{1}{121}+\dfrac{1}{122}+\dfrac{1}{123}+...+\dfrac{1}{150}\right)+\left(\dfrac{1}{151}+\dfrac{1}{152}+\dfrac{1}{153}+...+\dfrac{1}{180}\right)+\left(\dfrac{1}{181}+\dfrac{1}{182}+\dfrac{1}{183}+...+\dfrac{1}{200}\right)\)

\(\Leftrightarrow C>20\cdot\dfrac{1}{120}+30\cdot\dfrac{1}{150}+30\cdot\dfrac{1}{180}+20\cdot\dfrac{1}{200}\)

\(\Leftrightarrow C>\dfrac{1}{6}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{10}=\dfrac{19}{30}=\dfrac{76}{120}\)

\(\Leftrightarrow C>\dfrac{75}{120}=\dfrac{5}{8}\)

hay \(C>\dfrac{5}{8}\)(đpcm)

Ta có: 𝐶=1101+1102+1103+...+1200C=1011​+1021​+1031​+...+2001​

=(1101+1102+...+1120)+(1121+1122+1123+...+1150)+(1151+1152+1153+...+1180)+(1181+1182+1183+...+1200)=(1011​+1021​+...+1201​)+(1211​+1221​+1231​+...+1501​)+(1511​+1521​+1531​+...+1801​)+(1811​+1821...

Ta có 

1/101 > 1/150

1/102> 1/150

...>1/150

1/150 = 1/150

=> 1/101 + 1/102 + .... + 1/150 > 1/150 +1/150+....+1/150(50 số hạng )= 1/3

ta thấy 

1/101>1/150

1/102>1/150

...

1/150=1/150

=> 1/101+1/102+...+1/150>1/150+....+1/150(50 sô hạng)

=>1/101+1/102+...+1/150>1/3

1/151>1/200+

1/152>1/200

....

1/200=1/200

=>1/151+1/152+...+1/200>1/200+...+1/200(50 sô hạng)

=> 1/151+1/152+...+1/200>1/4

=> 1/101+...+1/200>1/3+1/4=7/12

a )   Số lượng số của dãy số trên là : 

\(\left(200-101\right):1+1=100\) ( số ) 

Do \(100⋮2\)nên ta nhóm dãy số trên thành 2 nhóm như sau : 

\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}=\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{150}\right)+\left(\frac{1}{151}+\frac{1}{152}+...+\frac{1}{200}\right)\)

\(\frac{1}{101}>\frac{1}{150};\frac{1}{102}>\frac{1}{150};...;\frac{1}{149}>\frac{1}{150};\frac{1}{150}=\frac{1}{150}\)

\(\Rightarrow\frac{1}{101}+\frac{1}{102}+...+\frac{1}{150}>\frac{1}{150}.50=\frac{1}{3}\left(1\right)\)

\(\frac{1}{151}>\frac{1}{200};\frac{1}{152}>\frac{1}{200};...;\frac{1}{199}>\frac{1}{200};\frac{1}{200}=\frac{1}{200}\)

\(\Rightarrow\frac{1}{151}+\frac{1}{152}+...+\frac{1}{200}>\frac{1}{200}.50=\frac{1}{4}\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\)

\(\Rightarrow\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}>\frac{1}{3}+\frac{1}{4}=\frac{7}{2}\left(3\right)\)

\(\frac{1}{101}< \frac{1}{100};\frac{1}{102}< \frac{1}{100};...;\frac{1}{199}< \frac{1}{100};\frac{1}{200}< \frac{1}{100}\)

\(\Rightarrow\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}< \frac{1}{100}.100=1\left(4\right)\)

Từ \(\left(3\right);\left(4\right)\Rightarrowđpcm\)

b )  Số lượng số dãy số trên là : 

\(\left(150-101\right):1+1=50\)( số ) 

Ta có : \(\frac{1}{101}>\frac{1}{150};\frac{1}{102}>\frac{1}{150};\frac{1}{103}>\frac{1}{150};...;\frac{1}{150}=\frac{1}{150}\)

\(\Rightarrow\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{150}>\frac{1}{150}.50=\frac{1}{3}\)

\(\Rightarrowđpcm\)

Ta có : \(\frac{1}{101}\) > \(\frac{1}{150}\)

            \(\frac{1}{102}\) > \(\frac{1}{150}\)

 .....................................................

             \(\frac{1}{149}\) > \(\frac{1}{150}\)

=> \(\frac{1}{101}\) + \(\frac{1}{102}\) + .......... + \(\frac{1}{150}\) > \(\frac{1}{150}\) + \(\frac{1}{150}\) + .......... +  \(\frac{1}{150}\)( có 50 p/s ) = \(\frac{1}{150}\) . 50 = \(\frac{1}{3}\)(1)

Ta lại có : \(\frac{1}{151}\) > \(\frac{1}{200}\)

                \(\frac{1}{152}\) > \(\frac{1}{200}\)

   ............................................

                 \(\frac{1}{199}\)> \(\frac{1}{200}\)

=> \(\frac{1}{151}\) + \(\frac{1}{152}\) + .................. + \(\frac{1}{200}\) > \(\frac{1}{200}\)+ \(\frac{1}{200}\) + ...................+ \(\frac{1}{200}\)(có 50 p/ )=\(\frac{1}{200}\) . 50 = \(\frac{1}{4}\)(2)

Từ (1) và (2) 

=> \(\frac{1}{101}\)+ \(\frac{1}{102}\) + \(\frac{1}{103}\) + ...................+ \(\frac{1}{200}\)>  \(\frac{1}{3}\) + \(\frac{1}{4}\) = \(\frac{4}{12}\) + \(\frac{3}{12}\) = \(\frac{7}{12}\)

Vậy A > \(\frac{7}{12}\)