Cho \(P=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2019.2021}\)
chứng tỏ rằng P<1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
11 tháng 7 2017
\(A=\dfrac{2^2}{1.3}+\dfrac{3^2}{2.4}+\dfrac{4^2}{3.5}+\dfrac{5^2}{4.6}+\dfrac{6^2}{5.7}\)
\(A=\dfrac{2.2.3.3.4.4.5.5.6.6}{1.3.2.4.3.5.4.6.5.7}\)
\(A=\dfrac{2.3.4.5.6}{1.2.3.4.5}.\dfrac{2.3.4.5.6}{3.4.5.6.7}\)
\(A=\dfrac{6}{1}.\dfrac{2}{7}=\dfrac{12}{7}\)
\(B=\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)\left(1+\dfrac{1}{9.11}\right)\)
\(B=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.\dfrac{100}{99}\)
\(B=\dfrac{4.9.16.100}{3.8.15.99}\)
\(B=\dfrac{2.2.3.3.4.4.10.10}{1.3.2.4.3.5.9.11}\)
\(B=\dfrac{2.3.4.10}{1.2.3.9}.\dfrac{2.3.4.10}{3.4.5.11}\)
\(B=10.\dfrac{2}{11}=\dfrac{20}{11}\)
4 tháng 5 2022
\(\dfrac{2}{1.3}\) + \(\dfrac{2}{3.5}\) + ..... + \(\dfrac{2}{95.97}\)
= 1 - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) + .... + \(\dfrac{1}{95}\) - \(\dfrac{1}{97}\)
= \(1-\dfrac{1}{97}\)
= \(\dfrac{96}{97}\)
4 tháng 5 2022
\(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+...+\dfrac{2}{95\times97}\)
\(=\dfrac{2}{3}\left(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+...+\dfrac{1}{95\times97}\right)\)
\(=\dfrac{2}{3}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{95}-\dfrac{1}{97}\right)\)
\(=\dfrac{2}{3}\left(1-\dfrac{1}{97}\right)\)\(=\dfrac{2}{3}\times\dfrac{96}{97}\)\(=\dfrac{64}{97}\)
H
2
NV
Nguyễn Việt Lâm
Giáo viên
11 tháng 3 2023
\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2021.2023}\)
\(=\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+...+\dfrac{2023-2021}{2021.2023}\)
\(=\dfrac{3}{1.3}-\dfrac{1}{1.3}+\dfrac{5}{3.5}-\dfrac{3}{3.5}+...+\dfrac{2023}{2021.2023}-\dfrac{2021}{2021.2023}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2021}-\dfrac{1}{2023}\)
\(=1-\dfrac{1}{2023}=\dfrac{2022}{2023}\)
H9
HT.Phong (9A5)
CTVHS
11 tháng 3 2023
\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}...+\dfrac{2}{2021.2023}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{2021}-\dfrac{1}{2023}\)
\(=1-\dfrac{1}{2023}\)
\(=\dfrac{2023}{2023}-\dfrac{1}{2023}\)
\(=\dfrac{2022}{2023}\)
DA
30 tháng 4 2022
\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{2020.2022}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{2020}-\dfrac{1}{2022}\)
\(=1-\dfrac{1}{2022}\)
\(=\dfrac{2021}{2022}\)
NH
11 tháng 3 2023
\(B=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}+\dfrac{2}{99\cdot101}\\ B=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{101}\\ B=\dfrac{1}{1}-\dfrac{1}{101}\\ B=\dfrac{101}{101}-\dfrac{1}{101}\\ B=\dfrac{100}{101}\)
NH
11 tháng 3 2023
\(\dfrac{2}{1\cdot3}=\dfrac{1}{1}-\dfrac{1}{3}=\dfrac{3}{3}-\dfrac{1}{3}=\dfrac{2}{3}\)
\(\dfrac{2}{3\cdot5}=\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{5}{15}-\dfrac{3}{15}=\dfrac{2}{15}\)
\(\dfrac{2}{5\cdot7}=\dfrac{1}{5}-\dfrac{1}{7}=\dfrac{7}{35}-\dfrac{5}{35}=\dfrac{2}{35}\)
và cứ như thế đến số cuối
27 tháng 5 2022
\(S=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{29\cdot31}\\ =\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{29}-\dfrac{1}{31}\\ =\dfrac{1}{1}-\dfrac{1}{31}\\ =\dfrac{30}{31}\)
mà \(\dfrac{30}{31}>\dfrac{2014}{2015}\Rightarrow S>P\)
NA
27 tháng 5 2022
So sánh vs j nhỉ .-.?
`S=1-1/3+1/3-1/5+...+1/29-1/31`
`S=1-1/31=30/31`
TD
18
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
15 tháng 2 2023
A = \(\dfrac{2}{1.3}\) + \(\dfrac{2}{3.5}\) + \(\dfrac{2}{5.7}\) + ...........+ \(\dfrac{2}{99.101}\)
A = \(\dfrac{1}{1}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\) +............+ \(\dfrac{1}{99}\) - \(\dfrac{1}{101}\)
A = \(\dfrac{1}{1}\) - \(\dfrac{1}{101}\)
A = \(\dfrac{100}{101}\)
18 tháng 5 2022
\(S=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{29}-\dfrac{1}{31}=1-\dfrac{1}{31}=\dfrac{30}{31}\)
P=2014/2015=1-1/2015
mà 1/31>1/2015
nên S<P
Ta có : \(P=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2019}-\dfrac{1}{2020}=1-\dfrac{1}{2020}=\dfrac{2019}{2020}\)
mà \(2019< 2020\)nên P < 1 ( đpcm )
\(P=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2019.2021}\)
\(P=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2019}-\dfrac{1}{2021}\)
\(P=1-\dfrac{1}{2021}\)
\(P=\dfrac{2020}{2021}\)
Vì \(\dfrac{2020}{2021}< 1\) ⇒ \(P< 1\) ( điều phải chứng minh )