Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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 > \(\dfrac{1}{2}\)
Easy!!
\(S=\dfrac{1}{21}+\dfrac{1}{22}+...+\dfrac{1}{35}>\dfrac{1}{29}+\dfrac{1}{29}+...+\dfrac{1}{29}\) (15 phân số \(\dfrac{1}{29}\))
\(=\dfrac{1.15}{29}=\dfrac{15}{29}>\dfrac{1}{2}\) (*)
\(\Rightarrow\dfrac{1}{21}+\dfrac{1}{22}+...+\dfrac{1}{35}>\dfrac{1}{2}^{\left(đpcm\right)}\)
P/s: đpcm là điều phải chứng minh
Có \(S=\dfrac{1}{21}+\dfrac{1}{22}+......+\dfrac{1}{35}\)
\(S=\dfrac{1}{21}+\dfrac{1}{22}+.........+\dfrac{1}{35}>\dfrac{1}{29}+\dfrac{1}{29}+\dfrac{1}{29}+........+\dfrac{1}{29}\)( 15 phân số \(\dfrac{1}{29}\))
\(S=\dfrac{15}{29}>\dfrac{1}{2}\)
\(S>\dfrac{1}{2}\)
Vậy S > \(\dfrac{1}{2}\)(đpcm)
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
đề sai hả bạn số hạng cuối có phải là \(\frac{1}{100}\)đúng không
ta co 1/50 >1/100
1/51>1/100
1/52>1/100
.........
1/99>1/100
suy ra S=1/50 +1/51 +1/52 +.....+1/99>1/100*50=1/2 suy ra S>1/2
https://www.youtube.com/watch?v=fBjsHQKClNA&index=7&list=PLq0mRSDfY0BAMTu98fNHi-Lg_E9BWDYhV
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