Cho :
A = 1.2 .3 ....2017.2018 . ( 1 + \(\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{2017}+\dfrac{1}{2018}\) )
CM : A\(\in\) N
A \(⋮\) 2019
K
Khách
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Những câu hỏi liên quan
17 tháng 1 2021
\(A=1.2.3...2018\left[\left(1+\dfrac{1}{2018}\right)+\left(\dfrac{1}{2}+\dfrac{1}{2017}\right)+...+\left(\dfrac{1}{1009}+\dfrac{1}{1010}\right)\right]\)
\(A=1.2.3...2018.2019\left(\dfrac{1}{1.2018}+\dfrac{1}{2.2017}+...+\dfrac{1}{1009.1010}\right)\)
\(\dfrac{A}{2019}=1.2.3...2018\left(\dfrac{1}{1.2018}+\dfrac{1}{2.2017}+...+\dfrac{1}{1009.1010}\right)\).
Rõ ràng tích 1 . 2 ... 2018 chia hết cho các tích 1 . 2018; 2 . 2017; ...; 1009 . 1010; do đó \(\dfrac{A}{2019}\) là số tự nhiên.
Vậy A chia hết cho 2019.
11 tháng 5 2023
\(A>\dfrac{2^{2018}}{2^{2018}+3^{2019}+5^{2020}}+\dfrac{3^{2019}}{2^{2018}+3^{2019}+5^{2020}}+\dfrac{5^{2020}}{5^{2020}+2^{2018}+3^{2019}}=1\)
\(B< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2019\cdot2020}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2019}-\dfrac{1}{2020}\)
=>B<1
=>A>B
MC
29 tháng 1 2019
Theo bài ra, ta có: \(B=\dfrac{2018}{1}+\dfrac{2017}{2}+\dfrac{2016}{3}+...+\dfrac{1}{2018}\)
\(B=\left(\dfrac{2018}{1}+1\right)+\left(\dfrac{2017}{2}+1\right)+\left(\dfrac{2016}{3}+1\right)+...+\left(\dfrac{1}{2018}+1\right)-2018\)
\(B=2019+\dfrac{2019}{2}+\dfrac{2019}{3}+...+\dfrac{2019}{2018}-2018\)
\(B=\dfrac{2019}{2}+\dfrac{2019}{3}+...+\dfrac{2019}{2018}+\left(2019-2018\right)\)
\(B=\dfrac{2019}{2}+\dfrac{2019}{3}+...+\dfrac{2019}{2018}+1\)
\(B=\dfrac{2019}{2}+\dfrac{2019}{3}+...+\dfrac{2019}{2018}+\dfrac{2019}{2019}\)
\(B=2019\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2019}\right)\)
Khi đó:\(\dfrac{B}{A}=\dfrac{2019\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2019}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2019}}\)
\(\Rightarrow\dfrac{B}{A}=2019\), là 1 số nguyên.
Vậy \(\dfrac{B}{A}\) là số nguyên.
HN
31 tháng 7 2018
a/ Ta có:
\(\dfrac{1}{\sqrt{n+1}+\sqrt{n}}=\dfrac{\left(\sqrt{n+1}-\sqrt{n}\right)}{\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}=\sqrt{n+1}-\sqrt{n}\)
\(\Rightarrow A=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{2019}-\sqrt{2018}=\sqrt{2019}-1\)
31 tháng 7 2018
a.\(A=\dfrac{1}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+\dfrac{1}{\sqrt{4}+\sqrt{3}}+...+\dfrac{1}{\sqrt{2019}+\sqrt{2018}}=\dfrac{\sqrt{2}-1}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}+\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}+...+\dfrac{\sqrt{2019}-\sqrt{2018}}{\left(\sqrt{2019}+\sqrt{2018}\right)\left(\sqrt{2019}-\sqrt{2018}\right)}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{2019}-\sqrt{2018}=\sqrt{2019}-1\)
10 tháng 6 2023
Ta biến đổi \(A=\dfrac{2-1}{1.2}+\dfrac{4-3}{3.4}+...+\dfrac{2016-2015}{2016.2015}+\dfrac{2018-2017}{2017.2018}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2015}-\dfrac{1}{2016}+\dfrac{1}{2017}-\dfrac{1}{2018}\)
\(A=\left(1+\dfrac{1}{3}+...+\dfrac{1}{2017}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2018}\right)\)
\(A=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2017}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2018}\right)\)
\(A=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2017}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{1009}\right)\)
\(A=\dfrac{1}{1010}+\dfrac{1}{1011}+...+\dfrac{1}{2017}+\dfrac{1}{2018}\)
Lại có \(B=\dfrac{1}{1010.2018}+\dfrac{1}{1011.2017}+...+\dfrac{1}{2018.1010}\)
\(B=\dfrac{1}{3028}.\left(\dfrac{3028}{1010.2018}+\dfrac{3028}{1011.2017}+...+\dfrac{3028}{2018.1010}\right)\)
\(B=\dfrac{1}{3028}\left(\dfrac{1}{1010}+\dfrac{1}{2018}+\dfrac{1}{1011}+\dfrac{1}{2017}+...+\dfrac{1}{2018}+\dfrac{1}{1010}\right)\)
\(B=\dfrac{1}{3028}.2\left(\dfrac{1}{1010}+\dfrac{1}{1011}+...+\dfrac{1}{2018}\right)\)
\(B=\dfrac{1}{3028}.2A\) \(\Rightarrow\dfrac{A}{B}=1514\inℤ\). Ta có đpcm
7 tháng 10 2021
a) \(\dfrac{1}{2!}+\dfrac{2}{3!}+...+\dfrac{2018}{2019!}\\ =\left(\dfrac{1}{1!}-\dfrac{1}{2!}\right)+\left(\dfrac{1}{2!}-\dfrac{1}{3!}\right)+...+\left(\dfrac{1}{2018!}-\dfrac{1}{2019!}\right)\\ =1-\dfrac{1}{2019!}< 1\)
7 tháng 10 2021
b) \(\dfrac{1\cdot2-1}{2!}+\dfrac{2\cdot3-1}{3!}+...+\dfrac{999\cdot1000-1}{1000!}\\ =\dfrac{1\cdot2}{2!}-\dfrac{1}{2!}+\dfrac{2\cdot3}{3!}-\dfrac{1}{3!}+...+\dfrac{999-1000}{1000!}-\dfrac{1}{1000!}\\ =\dfrac{1}{1!}-\dfrac{1}{2!}+\dfrac{1}{1!}-\dfrac{1}{3!}+\dfrac{1}{2!}-\dfrac{1}{4!}+...+\dfrac{1}{999!}+\dfrac{1}{1000!}\\ =1+1-\dfrac{1}{1000!}\\ =2-\dfrac{1}{1000!}< 2\)
LG
6 tháng 8 2017
Ta có : \(\dfrac{2017+2018}{2018+2019}=\dfrac{2017}{2018+2019}+\dfrac{2018}{2018+2019}\)
Rõ ràng ta thấy : \(\dfrac{2017}{2018}>\dfrac{2017}{2018+2019}\) (1)
\(\dfrac{2018}{2019}>\dfrac{2018}{2018+2019}\) (2)
Từ (1) và (2), suy ra :
\(\dfrac{2017}{2018}+\dfrac{2018}{2019}>\dfrac{2017+2018}{2018+2019}\)
Vậy ......................
~ Học tốt ~
LG
6 tháng 8 2017
Ta có : \(\dfrac{2017}{2018}+\dfrac{2018}{2019}+\dfrac{2019}{2020}=\left(1-\dfrac{1}{2018}\right)+\left(1-\dfrac{1}{2019}\right)+\left(1-\dfrac{1}{2020}\right)\)\(=\left(1+1+1\right)-\left(\dfrac{1}{2018}+\dfrac{1}{2019}+\dfrac{1}{2020}\right)\)
\(=3+\left(\dfrac{1}{2018}+\dfrac{1}{2019}+\dfrac{1}{2020}\right)< 3\)
Vậy \(\dfrac{2017}{2018}+\dfrac{2018}{2019}+\dfrac{2019}{2020}< 3\)