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 \(\dfrac{2017}{2018}+\dfrac{2018}{2019}+\dfrac{2019}{2020}< 3\)

a: \(\frac13\times\frac{2019}{2020}=\frac{2019}{3}\times\frac{1}{2020}=\frac{673}{2020}\)

b: \(2017\times\frac{2018}{2019}\times\frac{2019}{2018}\)

\(=2017\times\frac{2018}{2018}\times\frac{2019}{2019}\)

=2017x1x1

=2017

Ta có công thức tổng quát:

\(\frac{1}{\left(k+1\right)\cdot\sqrt{k}+k\cdot\sqrt{k+1}}\)

\(=\frac{1}{\sqrt{k\left(k+1\right)}\left(\sqrt{k+1}+\sqrt{k}\right)}=\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}}=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\)

\(\frac{1}{2+\sqrt2}+\frac{1}{3\sqrt2+2\sqrt3}+\cdots+\frac{1}{2019\sqrt{2018}+2018\sqrt{2019}}\)

\(=1-\frac{1}{\sqrt2}+\frac{1}{\sqrt2}-\frac{1}{\sqrt3}+\cdots+\frac{1}{\sqrt{2018}}-\frac{1}{\sqrt{2019}}\)

\(=1-\frac{1}{\sqrt{2019}}=\frac{\sqrt{2019}-1}{\sqrt{2019}}\)