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Ta có: 1/12>1/22 ; 1/13> 1/22.....1/21>1/22
Vậy: 1/12+1/13+...+1/22 > 1/22+1/22+1/22+...+1/22 = 11/22 = 1/2 (có 11 số hạng1/22).
hay: A>1/2
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 > \(\dfrac{1}{2}\)
Ta có S = \(\frac{1}{50}+\frac{1}{51}+\frac{1}{52}+...+\frac{1}{74}+\frac{1}{75}+\frac{1}{76}+\frac{1}{77}+...+\frac{1}{99}\)
\(=\left(\frac{1}{50}+\frac{1}{51}+\frac{1}{52}+...+\frac{1}{74}\right)+\left(\frac{1}{75}+\frac{1}{76}+\frac{1}{77}+...+\frac{1}{99}\right)\)
25 số hạng 25 số hạng
\(>\left(\frac{1}{75}+\frac{1}{75}+...+\frac{1}{75}\right)+\left(\frac{1}{100}+\frac{1}{100}+....+\frac{1}{100}\right)\)
\(=25.\frac{1}{75}+25.\frac{1}{100}=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}>\frac{6}{12}=\frac{1}{2}\)(ĐPCM)
Vậy S > 1/2
ta có:1/50>1/100
1/51>1/100
...............
1/99>1/100
=>S>50*1/100
=>S>1/2(đpcm)
1/50>1/100
1/51>1/100
...................
1/99>1/100
=>S>50*1/100(do từ 1/50 đến 1/99 có 50 số hạng)
=>S>1/2
\(A=\frac{10}{27}+\frac{9}{16}\frac{11}{34}\)
Ta có: \(\frac{10}{27}< >\backslash\left(\frac{9}{16}< >\backslash\left(\frac{11}{34}< >Nên\backslash\left(A< >b\right)\right)\right)\backslash\left(B=\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+...+\frac{1}{22}\right)\)
\(B>\frac{1}{22}+\frac{1}{22}+\frac{1}{22}+...+\frac{1}{22}=11.\frac{1}{22}=\frac{1}{2}\)
Nên \(B>\frac{1}{2}\)