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

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

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

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

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

hay \(C>\dfrac{5}{8}\)(đpcm)

Ta có:

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

\(c=\)(\(\frac{1}{101}\)\(+\)\(\frac{1}{102}\)\(+\)...\(+\)\(\frac{1}{120}\))\(+\)(\(\frac{1}{121}\)\(+\)\(\frac{1}{122}\)\(+\)...\(+\)\(\frac{1}{150}\))\(+\)(\(\frac{1}{151}\)\(+\)\(\frac{1}{152}\)\(+\)...\(+\)\(\frac{1}{180}\))\(+\)(\(\frac{1}{181}\)\(+\)\(\frac{1}{182}\)\(+\)...\(+\)\(\frac{1}{200}\))>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 !

ai tick mk với nào 

bn nhấn vào đúng 0 sẽ ra đáp án

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

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

\(\Rightarrowđpcm\)

a)             \(A=\frac{1}{5}+\frac{1}{45}+\frac{1}{117}+\frac{1}{221}+\frac{1}{357}+\frac{1}{525}\)

                  \(\Rightarrow A=\frac{1}{1.5}+\frac{1}{5.9}+\frac{1}{9.13}+...+\frac{1}{21.25}\)

                      \(\Rightarrow4A=\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{21.25}\)

                            \(4A=\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{21}-\frac{1}{25}\)

                             \(4A=\frac{1}{1}-\frac{1}{25}=\frac{24}{25}\)

                            \(\Rightarrow A=\frac{24}{25}\div4=\frac{6}{25}

\(A=\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+....+\dfrac{1}{200}\)

\(A=\sum\limits^{200}_{x=101}\left(\dfrac{1}{x+1}\right)=0,6857275648\)

Có: \(\dfrac{5}{8}=0.625\)

mà \(0,685...>0,625\)

\(\Rightarrow A>\dfrac{5}{8}\)

p/s: đây chỉ là 1 cách thoy, có cần lm cách khác k?

Lời Giải

Hay sử lý các con số khi không cần máy tính

\(A=\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{200}\)

dãy A có 100 số hạng \(⋮4=25\)

\(A=\left(\dfrac{1}{101}+...+\dfrac{1}{104}\right)+\left(\dfrac{1}{105}+..+\dfrac{1}{108}\right)+..+\left(\dfrac{1}{197}+\dfrac{1}{200}\right)\) Bao gồm (..)

\(A>B=\left(\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\right)\)

dãy A có 25 số hạng \(⋮5=5\)

\(B=\left(\dfrac{1}{26}+...+\dfrac{1}{30}\right)+..+\left(\dfrac{1}{46}+..+\dfrac{1}{50}\right)\)

\(B>C=\left(\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}\right)\)

\(\left\{{}\begin{matrix}\dfrac{1}{6}+\dfrac{1}{10}=\dfrac{16}{60}>\dfrac{16}{64}>\dfrac{2}{8}\\\dfrac{1}{7}+\dfrac{1}{9}=\dfrac{16}{63}>\dfrac{16}{64}>\dfrac{2}{8}\end{matrix}\right.\) \(\Rightarrow C>\dfrac{2}{8}+\dfrac{1}{8}+\dfrac{2}{8}=\dfrac{5}{8}\)

\(A>B>C>\dfrac{5}{8}\Rightarrow A>\dfrac{5}{8}\Rightarrow dpcm\Leftrightarrow dccm\)