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Ta có :
B = \(\dfrac{1}{2020}+\dfrac{2}{2019}+\dfrac{3}{2018}+...+\dfrac{2019}{2}+\dfrac{2020}{1}\)
B = \(\left(\dfrac{1}{2020}+1\right)+\left(\dfrac{2}{2019}+1\right)+\left(\dfrac{3}{2018}+1\right)+...+\left(\dfrac{2019}{2}+1\right)+1\)
B = \(\dfrac{2021}{2020}+\dfrac{2021}{2019}+\dfrac{2021}{2018}+...+\dfrac{2021}{2}+1\)
B = \(2021\left(\dfrac{1}{2021}+\dfrac{1}{2020}+\dfrac{1}{2019}+...+\dfrac{1}{2}\right)\) (1)
Mà A = \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}\) (2)
Từ (1) và (2) \(\Rightarrow\) \(\dfrac{A}{B}=\dfrac{1}{2021}\)
Ta có: \(B=\dfrac{1}{2020}+\dfrac{2}{2019}+\dfrac{3}{2018}+...+\dfrac{2019}{2}+\dfrac{2020}{1}\)
\(=\left(\dfrac{1}{2020}+1\right)+\left(\dfrac{2}{2019}+1\right)+\left(\dfrac{3}{2018}+1\right)+...+\left(\dfrac{2019}{2}+1\right)+1\)
\(=\dfrac{2021}{2020}+\dfrac{2021}{2019}+\dfrac{2021}{2018}+...+\dfrac{2021}{2}+\dfrac{2021}{2021}\)
Suy ra: \(\dfrac{A}{B}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}}{2021\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2021}\right)}=\dfrac{1}{2021}\)
\(B=\left(\dfrac{2020}{2}+1\right)+\left(\dfrac{2019}{3}+1\right)+...+\left(\dfrac{1}{2021}+1\right)+1\)
\(=\dfrac{2022}{2}+\dfrac{2022}{3}+...+\dfrac{2022}{2021}+\dfrac{2022}{2022}\)
=2022(1/2+1/3+...+1/2021+1/2022)
=>B/A=2022
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 ~
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 \(\dfrac{2017}{2018}+\dfrac{2018}{2019}+\dfrac{2019}{2020}< 3\)
Không tính thì sao mà làm được :)
a)
\(2020-\dfrac{1}{3^2}-\dfrac{1}{4^2}-...-\dfrac{1}{2019^2}\)
\(=3+\left(1-\dfrac{1}{3^2}\right)+\left(1-\dfrac{1}{4^2}\right)+....+\left(1-\dfrac{1}{2019^2}\right)\)
\(=3+\left(\dfrac{3^2-1}{3^2}+\dfrac{4^2-1}{4^2}+...+\dfrac{2019^2-1}{2019^2}\right)\)
\(=3+\left(\dfrac{2\cdot4}{3^2}+\dfrac{3\cdot5}{4^2}+\dfrac{4\cdot6}{5^2}+\dfrac{5\cdot7}{6^2}+...+\dfrac{2018\cdot2020}{2019^2}\right)\)
\(=3+\dfrac{\left(2\cdot3\cdot4\cdot....\cdot2018\right)}{3\cdot4\cdot5\cdot6...\cdot2019}\cdot\dfrac{\left(3\cdot4\cdot5\cdot....\cdot2020\right)}{3\cdot4\cdot5\cdot6\cdot....\cdot2019}=3+\dfrac{2\cdot2020}{2019}\)
\(=\dfrac{10097}{2019}\)
Có: \(\dfrac{1}{k^2}=\dfrac{1}{k.k}< \dfrac{1}{\left(k-1\right)k}\left(k\in\text{ℕ},k>0\right)\)
\(\Rightarrow A=2020-\dfrac{1}{3^2}-\dfrac{1}{4^2}-\dfrac{1}{5^2}-...-\dfrac{1}{2019^2}\)
\(A=2020-\left(\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+...+\dfrac{1}{2019^2}\right)\)
\(>2020-\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2018.2019}\right)\)
Có: \(\dfrac{1}{k-1}-\dfrac{1}{k}=\dfrac{1}{k\left(k-1\right)}\left(k\in\text{ℕ},k>0\right)\)
\(\Rightarrow A>2020-\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-...+\dfrac{1}{2018}-\dfrac{1}{2019}\right)\)
\(A>2020-\dfrac{1}{2}+\dfrac{1}{2019}\)>2,2
Có: \(B=\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{17}\)
\(B=\dfrac{1}{5}+\left(\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}\right)\)\(< \dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{6}+...+\dfrac{1}{6}\)
\(=\dfrac{1}{5}+\dfrac{1}{6}.12=2+\dfrac{1}{5}=2,2\)
Vậy A>B.
\(2021-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}\right)\)
\(=\left(1-1\right)+\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{3}\right)+...+\left(1-\dfrac{1}{2021}\right)\)
\(=\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{2020}{2021}\)
Giải:
\(2021-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}\right)=\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{2020}{2021}\)
Ta có:
\(2021-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}\right)\)
\(=\left(1-1\right)+\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{3}\right)+...+\left(1-\dfrac{1}{2021}\right)\)
\(=0+\dfrac{1}{2}+\dfrac{2}{3}+...+\dfrac{2020}{2021}\)
\(=\dfrac{1}{2}+\dfrac{2}{3}+...+\dfrac{2020}{2021}\)
Mà \(\dfrac{1}{2}+\dfrac{2}{3}+...+\dfrac{2020}{2021}=\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{2020}{2021}\)
\(\Rightarrow2021-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}\right)=\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{2020}{2021}\left(đpcm\right)\)
\(2T=2+\dfrac{3}{2^1}+\dfrac{4}{2^2}+...+\dfrac{2020}{2^{2018}}+\dfrac{2021}{2^{2019}}\)
\(T=2T-T=2+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2019}}-\dfrac{2021}{2^{2020}}\).
Đặt \(S=\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2019}}\Rightarrow2S=1+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2018}}\Rightarrow S=2S-S=1-\dfrac{1}{2^{2019}}\).
Từ đó \(T=2+1-\dfrac{1}{2^{2019}}-\dfrac{2021}{2^{2020}}< 3\).
Ta có :
\(\dfrac{2017}{2018}+\dfrac{2018}{2019}+\dfrac{2019}{2020}=\left(1-\dfrac{1}{2018}\right)+\left(1-\dfrac{1}{2019}\right)+\left(1-\dfrac{1}{2020}\right)\)
\(=\left(1+1+1\right)-\left(\dfrac{1}{2018}+\dfrac{1}{2019}+\dfrac{1}{2020}\right)\)
\(=3-\left(\dfrac{1}{2018}+\dfrac{1}{2019}+\dfrac{1}{2020}\right)< 3\)
\(\Leftrightarrow\dfrac{2017}{2018}+\dfrac{2018}{2019}+\dfrac{2019}{2020}< 3\)
Lời giải:
$A=1-\frac{1}{2019}+1-\frac{1}{2020}+1-\frac{1}{2021}+1+\frac{3}{2018}$
$=4+(\frac{1}{2018}-\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2020}+\frac{1}{2018}-\frac{1}{2021})$
$> 4+0+0+0+0=4$
Sửa đề: 2020/1+2019/2+...+1/2020
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}}{\left(1+\dfrac{2019}{2}\right)+\left(1+\dfrac{2018}{3}\right)+...+\dfrac{1}{2020}+1+1}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}}{\dfrac{2021}{2}+\dfrac{2021}{3}+...+\dfrac{2021}{2020}+\dfrac{2021}{2021}}\)
=1/2021