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Ta có 1+5/28=33/28
Đặt A=1/11+1/12+1/13+...+1/69+1/70
A=(1/11+1/12++1/13+...+1/20)+(1/21+1/22+1/23+...+1/30)+(1/31+1/32+1/33+...1/60)+...+1/70
Ta thấy :
1/11+1/12+1/13+...+1/20>1/20+1/20+1/20+...+1/20(có 10 số hạng 1/20)=1/20*10=1/2
1/21+1/22+1/23+...+1/30>1/30+1/30+1/30+...+1/30(10 số hạng 1/30)=1/30*10=1/3
1/30+1/31+1/32+...+1/60>1/60+1/60+...+1/60(30 số hạng 1/60)=1/60*30=1/2
1/61+1/62+1/63+...+1/70>1/70+1/70+1/70+...+1/70(10 số hạng 1/70)=1/70*10=1/7
=>1/11+1/12+1/13+...+1/69+1/70>1/2+1/3+1/2+1/7
=>A>31/21
Mà 31/21>33/28
=>A>33/28
=>A>1+5/28(DPCM)
Vậy A>1+5/28
a, Ta có: \(A=\frac{1}{11}+\frac{1}{12}+...+\frac{1}{50}=\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{30}\right)+\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}\right)\)
Nhận xét: \(\frac{1}{11}+\frac{1}{12}+....+\frac{1}{30}>\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}=\frac{20}{30}=\frac{2}{3}\)
\(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}=\frac{20}{60}=\frac{1}{3}\)
\(\Rightarrow A>\frac{2}{3}+\frac{1}{3}=1>\frac{1}{2}\)
Vậy A > 1/2
b, Ta có: \(\frac{1}{50}>\frac{1}{100};\frac{1}{51}>\frac{1}{100};........;\frac{1}{99}>\frac{1}{100}\)
\(\Rightarrow B>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{50}{100}=\frac{1}{2}\)
Vậy B > 1/2
c, Ta có: \(C=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}=\frac{1}{10}+\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}\right)\)
Nhận xét: \(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{90}{100}=\frac{9}{10}\)
\(\Rightarrow C>\frac{1}{10}+\frac{9}{10}=\frac{10}{10}=1\)
Vậy C > 1
a: Ta có
A = \(\dfrac{1}{10}\) + \((\dfrac{1}{11}\) + \(\dfrac{1}{12}\) + ...+ \(\dfrac{1}{100}\)\()\)
⇒ A > \(\dfrac{1}{10}\) + \((\dfrac{1}{100}\) + \(\dfrac{1}{100}\) + ...+ \(\dfrac{1}{100}\)\()\)90 số hạng
⇒ A > \(\dfrac{1}{10}\) + \(\dfrac{90}{100}\)
⇒ A > 1
vậy A > 1
b: ta có
S = (\(\dfrac{1}{21}\) + \(\dfrac{1}{22}\)+ \(\dfrac{1}{23}\) + \(\dfrac{1}{24}\) + \(\dfrac{1}{25}\))+(\(\dfrac{1}{26}\) + \(\dfrac{1}{27}\)+ \(\dfrac{1}{28}\) + \(\dfrac{1}{29}\) + \(\dfrac{1}{30}\))+(\(\dfrac{1}{31}\) + \(\dfrac{1}{32}\)+ \(\dfrac{1}{33}\) + \(\dfrac{1}{34}\) + \(\dfrac{1}{35}\))
⇒ S > (\(\dfrac{1}{25}\) + \(\dfrac{1}{25}\)+ \(\dfrac{1}{25}\) + \(\dfrac{1}{25}\) + \(\dfrac{1}{25}\))+(\(\dfrac{1}{30}\) + \(\dfrac{1}{30}\)+ \(\dfrac{1}{30}\) + \(\dfrac{1}{30}\) + \(\dfrac{1}{30}\))+(\(\dfrac{1}{35}\) + \(\dfrac{1}{35}\)+ \(\dfrac{1}{35}\) + \(\dfrac{1}{35}\) + \(\dfrac{1}{35}\))
⇔ S > \(\dfrac{5}{25}\)+\(\dfrac{5}{30}\)+\(\dfrac{5}{35}\)
⇔ S > \(\dfrac{1}{5}\)+\(\dfrac{1}{6}\)+\(\dfrac{1}{7}\)
⇔ S > \(\dfrac{107}{210}\)> \(\dfrac{105}{210}\)=\(\dfrac{1}{2}\)
vậy S > \(\dfrac{1}{2}\)
\(A = (\frac{1}{10} + ...+ \frac{1}{19} ) + (\frac{1}{20} + ...+ \frac{1}{29}) + (\frac{1}{30} +...+ \frac{1}{39} ) + (\frac{1}{40} + ...+\frac{1}{49} ) + (\frac{1}{50} +....+ \frac{1}{59}) + (\frac{1}{60} + ....+\frac{1}{69}) + \frac{1}{70}\)
Ta có : mỗi bên có 10 số hạng
\( (\frac{1}{10} + ..+ \frac{1}{19}) < (\frac{1}{10} + ...+ \frac{1}{10}) = \frac{1}{1}\)
\(\frac{1}{20}+..+ \frac{1}{29} < (\frac{1}{20}+..+\frac{1}{20}) = \frac{1}{2}\)
\((\frac{1}{30} +...+ \frac{1}{39} )< (\frac{1}{30} +...+ \frac{1}{30}) = \frac{1}{3}\)
\((\frac{1}{40} + ...+\frac{1}{49} )< (\frac{1}{40} + ...+\frac{1}{40}) = \frac{1}{4}\)
\((\frac{1}{50} +....+ \frac{1}{59})< (\frac{1}{50} +....+ \frac{1}{50}) = \frac{1}{5}\)
\((\frac{1}{60} + ....+\frac{1}{69}) + \frac{1}{70}< (\frac{1}{60} + ....+\frac{1}{60})+ \frac{1}{70} = \frac{1}{6} +\frac{1}{70}\)
\(\implies A < 1+\frac{1}{2} + ...+ \frac{1}{6} + \frac{1}{70}= \frac{13}{15} + \frac{1}{70} <1<\frac {51}{20} \)
\(\implies A<\frac{51}{20}\) \((đpcm)\)
Ko bt