j mà  nhìu zu zậy làm bao giờ mới xong

Ủng hộ mk đi các bạn
 

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​+1831​+...+2001​)

⇔𝐶>20⋅1120+30⋅1150+30⋅1180+20⋅1200⇔C>20⋅1201​+30⋅1501​+30⋅1801​+20⋅2001​

⇔𝐶>16+15+16+110=1930=76120⇔C>61​+51​+61​+101​=3019​=12076​

⇔𝐶>75120=58⇔C>12075​=85​

hay 𝐶>58C>85​(đpcm)

 TỰ thay C=a nhA

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

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

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

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

=> \(\frac{1}{101}\) + \(\frac{1}{102}\) + .......... + \(\frac{1}{150}\) > \(\frac{1}{150}\) + \(\frac{1}{150}\) + .......... +  \(\frac{1}{150}\)( 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 > \(\frac{7}{12}\)