Đặt \(A=\left(\sqrt{2018}+\sqrt{2020}\right)\)

\(\Rightarrow A^2=2018+2\sqrt{2018.2020}+2020=4038+\sqrt{4.2018.2020}=4038+\sqrt{4.\left(2019^2-1\right)}\)

Đặt \(B=2\sqrt{2019}=\sqrt{4.2019}\)

\(B^2=4.2019=2.2019+2.2019=4038+\sqrt{4.2019^2}\)

=> \(\sqrt{4.2019^2}>\sqrt{4.\left(2019^2-1\right)}\)

\(\Rightarrow A>B\Leftrightarrow\sqrt{2018}+\sqrt{2020}>2\sqrt{2019}\)

\(7^{2019}-7^{2020}=7^{2019}\left(1-7\right)\)

\(7^{2018}-7^{2019}=7^{2018}\left(1-7\right)\)

Mà \(7^{2019}>7^{2018}\)

\(\Rightarrow7^{2019}-7^{2020}>7^{2018}-7^{2019}\)

# Học tốt

\(7^{2019}-7^{2020}=7^{2019}-7\cdot7^{2019}=-6.7^{2019}\)  

\(7^{2018}-7^{2019}=7^{2018}-7\cdot7^{2018}=-6\cdot7^{2018}\)

vì \(7^{2019}>7^{2018}\Rightarrow-6\cdot7^{2019}< -6\cdot7^{2018}\)   

Vậy \(7^{2019}-7^{2020}< 7^{2018}-7^{2019}\)

Ta có: \(7^{2019}-7^{2020}=7^{2019}.\left(1-7\right)=\left(-6\right).7^{2019}\)

           \(7^{2018}-7^{2019}=7^{2018}.\left(1-7\right)=\left(-6\right).7^{2018}\)

Vì \(7^{2019}>7^{2018}\)\(\Rightarrow\)\(6.7^{2019}>6.7^{2018}\)\(\Rightarrow\)\(\left(-6\right).7^{2019}< \left(-6\right).7^{2018}\)

             \(\Rightarrow\)\(7^{2019}-7^{2020}< 7^{2018}-7^{2019}\)

\(8^2=64=32+2\sqrt{16^2}\)

\(\left(\sqrt{15}+\sqrt{17}\right)^2=32+2\sqrt{15.17}=32+2\sqrt{\left(16-1\right)\left(16+1\right)}\)

\(=32+2\sqrt{16^2-1}\)

\(< =>8^2>\left(\sqrt{15}+\sqrt{17}\right)^2\)

\(8>\sqrt{15}+\sqrt{17}\)

\(\left(\sqrt{2019}+\sqrt{2021}\right)^2=4040+2\sqrt{2019.2021}\)

\(=4040+2\sqrt{\left(2020-1\right)\left(2020+1\right)}=4040+2\sqrt{2020^2-1}\)

\(\left(2\sqrt{2020}\right)^2=8080=4040+2\sqrt{2020^2}\)

\(< =>\sqrt{2019}+\sqrt{2021}< 2\sqrt{2020}\)

mik chọn điền

< 

mik lười chép ại đề bài 

\(\dfrac{2019}{2020}=1-\dfrac{1}{2020}>1-\dfrac{1}{2019}=\dfrac{2018}{2019}\)

\(\dfrac{2019}{2020}>\dfrac{2018}{2019}\)

ta có : 

A = \(\dfrac{5^{2020}+1}{5^{2020}+1}\)

B = \(\dfrac{5^{2019}+1}{5^{2020}+1}\)

\(\Leftrightarrow\) B < A

Ta có : \(\hept{\begin{cases}\frac{2019}{2020}< 1\\\frac{2018}{2018}=1\end{cases}\Rightarrow\frac{2019}{2020}< \frac{2018}{2018}}\)

Ta có : 

\(\frac{2019}{2020}< 1\)

\(\frac{2018}{2018}=1\)

\(\Rightarrow\frac{2019}{2020}< \frac{2018}{2018}\)

#Riin

A = \(\dfrac{5^{2020}+1}{5^{2021}+1}\) ⇒ A \(\times\) 10 = 2 \(\times\)5 \(\times\) \(\dfrac{5^{2020}+1}{5^{2021}+1}\) =2\(\times\) \(\dfrac{5^{2021}+5}{5^{2021}+1}\)

10A =2 \(\times\) \(\dfrac{5^{2021}+5}{5^{2021}+1}\) = 2 \(\times\)(1 + \(\dfrac{4}{5^{2021}+1}\) )= 2 + \(\dfrac{8}{5^{2021}+1}\) >2

B = \(\dfrac{10^{2019}+1}{10^{2020}+1}\) ⇒ B \(\times\) 10 = 10 \(\times\) \(\dfrac{10^{2019}+1}{10^{2020}+1}\)= \(\dfrac{10^{2020}+10}{10^{2020}+1}\)

10B = \(\dfrac{10^{2020}+10}{10^{2020}+1}\) = 1 + \(\dfrac{9}{10^{2020}+1}\) < 2

10A > 2 > 10B ⇒ 10A>10B ⇒ A>B