so sánh 1/1x2+1/2x3+...+49/50 và 1
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)
1/1.2+1/2.3+1/3.4+...+1/59.60
=1-1/2+1/2-1/3+1/3-1/4+...+1/59-1/60
=1-1/60
=59/60
vì 1>59/60
=> 1>1/1.2+1/2.3+1/3.4+...+1/59.60
chúc bạn học tốt nha
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{59.60}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{59}-\frac{1}{60}\)
\(=1-\frac{1}{60}=\frac{59}{60}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
A = 1.2 + 2.3 + 3.4 + ... + 2017.2018
⇒ 3A = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 2017.218.(2019 - 2016)
= 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 2017.2018.2019 - 2016.2017.2018
= 2017.2018.2019
= 2017.2018.2019
B = 2018³/3 ⇒ 3B = 2018³
Ta có:
2017.2019 = (2018 - 1).(2018 + 1)
= 2018² - 1²
= 2018.2018 - 1 < 2018.2018
⇒ 2017.2018.2019 < 2018.2018.2018
⇒ 3A < 3B
⇒ A < B
![](https://rs.olm.vn/images/avt/0.png?1311)
S= 2x(1/1x2+1/2x3+1/3x4+...........+1/2020x2021)
S=2x(1-1/2+1/2-1/3+1/3-...+1/2020-1/2021)
S=2x(1-1/2021)
S=2x2020/2021
S=4040/2021
2019/2010<3/2<4040/2021
=>2019/2010<S
S = 2 x (\(\frac{2}{1\times2}+\frac{2}{2\times3}+\frac{2}{3\times4}+...+\)\(\frac{2}{2020\times2021}\))
= 2 x (\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\)\(\frac{1}{2020\times2021}\))
= 2 x ( \(1-\frac{1}{2021}\))
= \(2\times\frac{2020}{2021}\)
= \(\frac{4040}{2021}\)
= \(\frac{4042-2}{2021}\)
\(=2-\frac{2}{2021}\)
Ta có :
\(\frac{2019}{2010}=\frac{2020-1}{2010}=2-\frac{1}{2010}=2-\frac{2}{2020}\)
Ta thấy \(\frac{2}{2021}< \frac{2}{2020}\)
nên \(2-\frac{2}{2021}>2-\frac{2}{2020}\)
Vậy \(S\)\(>\frac{2019}{2010}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+....+\dfrac{1}{24\times25}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{24}-\dfrac{1}{25}\)
\(=1-\dfrac{1}{25}\)
\(=\dfrac{24}{25}\)
Trả lời
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{49\cdot50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}\)
\(\Rightarrow\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{49\cdot50}< 1\)
Vậy \(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{49\cdot50}< 1\left(đpcm\right)\)
1-1/50=49/50<1