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30 số hạng đầu lớn hơn 1
\(\frac{1}{10}+\frac{1}{11}+..+\frac{1}{19}>\frac{1}{20}+\frac{1}{20}+..+\frac{1}{20}=\frac{1}{2}\)\(\frac{1}{2}\)
\(\frac{1}{20}+\frac{1}{21}+..+\frac{1}{29}>\frac{1}{30}+\frac{1}{30}+..+\frac{1}{30}=\frac{1}{3}\)
\(\frac{1}{30}+\frac{1}{31}+...+\frac{1}{39}>\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}=\frac{1}{4}\)
=> \(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{39}>\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{13}{12}>1\)
\(A=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\)
\(A=\frac{1}{10}+\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\right)>\frac{1}{10}+\left(\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\right)\)
\(A=\frac{1}{10}+\frac{99}{100}=1\)
=> A > 1
\(A=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}+\frac{1}{100}\)
\(A=\frac{1}{10}+\frac{1}{11}+...+\frac{1}{19}>\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}=\frac{10}{20}=\frac{1}{2}\)
\(A=\frac{1}{20}+\frac{1}{21}+...+\frac{1}{29}>\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}=\frac{10}{30}=\frac{1}{3}\)
\(A=\frac{1}{30}+\frac{1}{31}+...+\frac{1}{39}>\frac{1}{40}+\frac{1}{40}+... +\frac{1}{40}=\frac{10}{40}=\frac{1}{4}\)
\(\Rightarrow A>1\)
\(C=\left(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)\)
\(>\frac{1}{50}.41+\frac{1}{100}.50=\frac{41}{50}+\frac{50}{100}=\frac{33}{25}=1\frac{8}{25}>1\)
Ta thấy rằng mỗi số hạng trong tổng đều lớn hơn hoặc bằng \(\frac{1}{100}\)
=> \(C>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{1}{100}x100=1\)
=> C>1 (Đpcm)
1/10+1/11+……+1/99 > 1/20+1/20+…..+1/20 = 10/20 = 1/2
1/20+1/21+……+1/29 > 1/20+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
Vậy A > 1
Ta có:
\(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{49}>\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}=\frac{40}{50}=\frac{4}{5}\)
\(\frac{1}{50}+\frac{1}{51}+...+\frac{1}{99}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{50}{100}=\frac{1}{2}\)
Từ đây ta suy ra
A > \(\frac{4}{5}+\frac{1}{2}+\frac{1}{100}=1,31>1\)
+)Ta có:\(A=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+..........+\frac{1}{99}+\frac{1}{100}\)(có (100-10):1+1=91 số hạng)
\(\Rightarrow A=\left(\frac{1}{10}+\frac{1}{11}+.........+\frac{1}{54}\right)+\frac{1}{55}+\left(\frac{1}{56}+\frac{1}{57}+.............+\frac{1}{100}\right)>\)
\(\left(\frac{1}{54}+\frac{1}{54}+........+\frac{1}{54}\right)+\frac{1}{55}+\left(\frac{1}{100}+\frac{1}{100}+........+\frac{1}{100}\right)\)
\(\Rightarrow A>\frac{45}{54}+\frac{1}{55}+\frac{45}{100}=\frac{5}{6}+\frac{1}{55}+\frac{9}{20}=\frac{5}{6}+\frac{9}{20}+\frac{1}{55}=\frac{50}{60}+\frac{27}{60}+\frac{1}{55}\)\(=\frac{77}{60}+\frac{1}{55}>1\)(vì \(\frac{77}{60}>1\))
\(\Rightarrow A>1\)(ĐPCM)
Chúc bn học tốt
Đặt A = B + \(\frac{1}{10}\) Ta thấy B có 90 số hạng và 1/100 < 1/11 ; 1/100 < 1/12 .....
Giả sử cả 90 số hạng đều là 1/100 ta có B > 90.(1/100) = 90/100
=> A > 1/10 + 90/100 => A>1
\(A=\frac{1}{10}+\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}\right)\)
\(A>\frac{1}{10}+\frac{1}{10}.90=1\)
Vậy A>1
Ta có : \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+....+\frac{1}{19}>\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}=\frac{1}{20}.10=\frac{1}{2}\) ( 10 số hạng 1/20)
\(\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+....+\frac{1}{29}>\frac{1}{30}+\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}=\frac{1}{30}.10=\frac{1}{3}\) ( 10 số hạng 1/30 )
.....................................
\(\frac{1}{90}+\frac{1}{91}+...+\frac{1}{99}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{1}{100}.10=\frac{1}{10}\). Và: \(\frac{1}{100}=\frac{1}{100}\)
Nên: \(C=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}>\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}+\frac{1}{100}>1\) (đpcm)
Ta có:
\(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{19}>\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}=\frac{10}{20}=\frac{1}{2}\)
\(\frac{1}{20}+\frac{1}{21}+...+\frac{1}{19}>\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}=\frac{10}{30}=\frac{1}{3}\)
\(\frac{1}{30}+\frac{1}{31}+...+\frac{1}{39}>\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}+\frac{10}{40}+\frac{1}{4}\)
\(=>\frac{1}{10}+\frac{1}{11}+...+\frac{1}{39}>\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{13}{12}>1\)
Vậy \(C>1\)
Làm ơn giải ra luôn hộ
Ta có: A = \(\frac{1}{10}+\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{99}\right)\)
Nhận xét: \(\frac{1}{11}>\frac{1}{100};\frac{1}{12}>\frac{1}{100};...;\frac{1}{99}>\frac{1}{100}\)
\(\Rightarrow A>\frac{1}{10}+\left(\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}\right)=\frac{1}{10}+\frac{90}{100}=1\)
Vậy A > 1 (đpcm)