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5 tháng 4

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 > \(\dfrac{1}{2}\)

 

14 tháng 5 2015

1/10+1/11+…+1/19 > 1/20+1/20+…+1/20 = 10/20 = 1/2 
1/20+1/21+…+1/29 > 1/30+1/30+…+1/30 = 10/30 = 1/3 
1/30+1/31+…+1/39 > 1/40+1/40+…+1/40 = 10/40 = 1/4 
=> A>1

14 tháng 5 2015

giangtuantai ơi ! Bạn vẫn đi copy à ?

A=12.34.56...99100

⇒A<23.45.67...100101

⇒A2<23.45.67...100101.12.34.56...99100

⇒A2<1101<1100=1102

A=12.34.56...99100

⇒A<23.45.67...100101

⇒A2<23.45.67...100101.12.34.56...99100

⇒A2<1101<1100=1102

14 tháng 3 2016

Chỉ cần 30 số hạng đầu đã lớn hơn 1. 
1/10+1/11+…+1/19 > 1/20+1/20+…+1/20 = 10/20 = 1/2 
1/20+1/21+…+1/29 > 1/30+1/30+…+1/30 = 10/30 = 1/3 
1/30+1/31+…+1/39 > 1/40+1/40+…+1/40 = 10/40 = 1/4 
=> 
1/10+1/11+…+1/39 > 1/2+1/3+1/4 = 13/12 > 1

9 tháng 3 2017

Ta có :

A = \(\dfrac{1}{10}\) + \(\dfrac{1}{11}\) + \(\dfrac{1}{12}\) +.................+ \(\dfrac{1}{99}\) + \(\dfrac{1}{100}\) ( 91 số hạng)

A = \(\dfrac{1}{10}\) + \(\left(\dfrac{1}{11}+\dfrac{1}{12}+...........+\dfrac{1}{99}+\dfrac{1}{100}\right)\)

\(\dfrac{1}{11}>\dfrac{1}{100}\)

\(\dfrac{1}{12}>\dfrac{1}{100}\)

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

\(\dfrac{1}{99}< \dfrac{1}{100}\)

\(=>\) \(A\) > \(\dfrac{1}{10}+\left(\dfrac{1}{100}+\dfrac{1}{100}+........+\dfrac{1}{100}\right)\) (90 số hạng \(\dfrac{1}{100}\) )

A > \(\dfrac{1}{10}+\dfrac{90}{100}\)

\(A\) > \(\dfrac{1}{10}+\dfrac{9}{10}\)

=> A > 1

=> đpcm

8 tháng 3 2017

\(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{19}>\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}=\frac{10}{20}=\frac{1}{2}\)

\(\frac{1}{20}+\frac{1}{21}+...+\frac{1}{29}>\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}=\frac{10}{30}=\frac{1}{3}\)

\(\frac{1}{30}+\frac{1}{31}+...+\frac{1}{39}>\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}=\frac{10}{40}=\frac{1}{4}\)

\(\Rightarrow\frac{1}{10}+\frac{1}{11}+...+\frac{1}{39}>\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\)

            \(\frac{13}{12}\)        \(>\)         \(1\)