![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
=> 2A =2 + 22 + 23 + ... + 22020
=> 2A-A =( 2 + 22 + 23 + ... + 22020)- (1 + 2 + 22 + 23 + ... + 22019)
=> A =22020-1
=> A+1 =22020
Vậy A + 1 là một số chính phương
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=1+2+2^2+...+2^{2018}\)
\(2A=2+2^3+2^4+...+2^{2019}\)
\(A=2A-A=1-2^{2019}\)
\(B-A=2^{2019}-\left(1-2^{2019}\right)\)
\(B-A=2^{2019}-1+2^{2019}\)
\(B-A=1\)
`#3107`
\(A=1+2+2^2+2^3+...+2^{2018}\) và \(B=2^{2019}\)
Ta có:
\(A=1+2+2^2+2^3+...+2^{2018}\)
\(2A=2+2^2+2^3+...+2^{2019}\)
\(2A-A=\left(2+2^2+2^3+...+2^{2019}\right)-\left(1+2+2^2+2^3+...+2^{2018}\right)\)
\(A=2+2^2+2^3+...+2^{2019}-1-2-2^2-2^3-...-2^{2018}\)
\(A=2^{2019}-1\)
Vậy, \(A=2^{2019}-1\)
Ta có:
\(B-A=2^{2019}-2^{2019}+1=1\)
Vậy, `B - A = 1.`
![](https://rs.olm.vn/images/avt/0.png?1311)
Sửa đề: \(S=\dfrac{1}{20}+\dfrac{1}{21}+\dfrac{1}{22}+...+\dfrac{1}{50}\)
Ta có: \(S=\dfrac{1}{20}+\dfrac{1}{21}+\dfrac{1}{22}+...+\dfrac{1}{50}\)
\(=\dfrac{1}{20}+\left(\dfrac{1}{21}+\dfrac{1}{22}+...+\dfrac{1}{30}\right)+\left(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}\right)+\left(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}\right)\)
\(\Leftrightarrow S>\dfrac{1}{20}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}=\dfrac{1}{4}+\dfrac{1}{3}+\dfrac{1}{4}\)
\(\Leftrightarrow S>\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{3}{4}\)(đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Giải
Đặt A=1/21+1/22+1/23+1/24+...+1/80
Ta có:
A=(1/21+1/22+...+1/40)+(1/41+...+1/80)
→A>(1/40+1/40+...+1/40)+(1/80+..+1/80)
→A>20/40+40/80
→A>1/2+1/2
→A>1 (1)
Lại có:
A=(1/21+1/22+...+1/40)+(1/41+...+1/80)
→A<(1/20+1/20+...+1/20)+(1/40+...+1/40)
→A<20/20+40/40
→A<2 (2)
Từ (1),(2)→1<A<2
→A không là số tự nhiên
Đặt A=1/21+1/22+1/23+1/24+...+1/80
Ta có:
A=(1/21+1/22+...+1/40)+(1/41+...+1/80)
→A>(1/40+1/40+...+1/40)+(1/80+..+1/80)
→A>20/40+40/80
→A>1/2+1/2
→A>1 (1)
Lại có:
A=(1/21+1/22+...+1/40)+(1/41+...+1/80)
→A<(1/20+1/20+...+1/20)+(1/40+...+1/40)
→A<20/20+40/40
→A<2 (2)
Từ (1),(2)→1<A<2
→A không là số tự nhiên
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=\frac{1}{21}+\frac{1}{22}+...+\frac{1}{150}\)
\(=\left(\frac{1}{21}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+...+\frac{1}{80}\right)+\left(\frac{1}{81}+...+\frac{1}{150}\right)\)
\(>\left(\frac{1}{40}+...+\frac{1}{40}\right)+\left(\frac{1}{80}+...+\frac{1}{80}\right)+\left(\frac{1}{150}+...+\frac{1}{150}\right)\)
\(=\frac{20}{40}+\frac{40}{80}+\frac{70}{150}\)
\(=\frac{1}{2}+\frac{1}{2}+\frac{7}{15}>\frac{5}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
do \(\frac{5}{20}< 1;\frac{5}{21}< 1;\frac{5}{22}< 1;\frac{5}{23}< 1;\frac{5}{24}< 1\)
\(\Rightarrow\frac{5}{20}+\frac{5}{21}+\frac{5}{22}+\frac{5}{23}+\frac{5}{24}< 1\)
Vậy S < 1
Mk nghĩ thế bn ạ
Ai thấy tớ đúng ủng hộ nha
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(\frac{5}{20}>\frac{5}{25}\)
\(\frac{5}{21}>\frac{5}{25}\)
\(\frac{5}{22}>\frac{5}{25}\)
\(\frac{5}{23}>\frac{5}{25}\)
\(\frac{5}{24}>\frac{5}{25}\)
\(\Rightarrow\frac{5}{20}+\frac{5}{21}+\frac{5}{22}+\frac{5}{23}+\frac{5}{24}>5.\frac{5}{25}=1\)
\(\Rightarrow\frac{5}{20}+\frac{5}{21}+\frac{5}{22}+\frac{5}{23}+\frac{5}{24}>1\)
ta có S=5/20+5/21+5/22+5/23+5/24>5/25+5/25+5/25+5/25+5/25=5/25*5=1
=>đpcm
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\dfrac{21}{22}+\dfrac{22}{23}=\dfrac{967}{506}>1\)
\(B=\dfrac{21+22}{22+23}=\dfrac{43}{45}< 1\)
Vậy \(A>B\)
\(\dfrac{21}{22}\) > \(\dfrac{21}{22+23}\)
\(\dfrac{22}{23}\) > \(\dfrac{22}{22+23}\)
Cộng vế với vế ta có:
A = \(\dfrac{21}{22}\) + \(\dfrac{22}{23}\) > \(\dfrac{21+22}{22+23}\) = B ⇒ A > B
\(S=1+2+2^2+2^3+...+2^{2019}\)
\(2S=2+2^2+2^3+2^4+...+2^{2020}\)
\(2S-S=\left(2+2^2+2^3+...+2^{2020}\right)-\left(1+2+2^2+...+2^{2019}\right)\)
\(S=2^{2020}-1\)
Cho hỏi đề bài yêu cầu gì?