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 \(B=2\sqrt{2019}=\sqrt{4.2019}\)

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

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

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

1,Ta có : \(\sqrt{11}-\sqrt{10}=\frac{11-10}{\sqrt{11}+\sqrt{10}}=\frac{1}{\sqrt{11}+\sqrt{10}}\)

\(\sqrt{6}-\sqrt{5}=\frac{6-5}{\sqrt{6}-\sqrt{5}}=\frac{1}{\sqrt{6}-\sqrt{5}}\)

Dễ thấy : \(11+10>6+5\Rightarrow\sqrt{11}+\sqrt{10}>\sqrt{6}+\sqrt{5}\)

từ đó suy ra : \(\frac{1}{\sqrt{11}+\sqrt{10}}< \frac{1}{\sqrt{6}+\sqrt{5}}\)( theo so sánh phân số có cùng tử )

Vậy...

2,\(\sqrt{2019}+\sqrt{2021}và2\sqrt{2020}\)

Giả sử : \(\sqrt{2019}+\sqrt{2021}< 2\sqrt{2020}\)

\(\Leftrightarrow\left(\sqrt{2019}+\sqrt{2021}\right)^2< \left(2\sqrt{2020}\right)^2\) ( bình phương 2 vế )

\(\Leftrightarrow2019+2021+2\sqrt{2019.2021}< 4.2020\)

\(\Leftrightarrow4040+2\sqrt{2020^2-1^2}< 8080\)

\(\Leftrightarrow\)\(4040+\left(-4040\right)+2\left|2020-1\right|< 8080+\left(-4040\right)\)

( cộng cả hai vế với -4040)

\(\Leftrightarrow2.2019< 4040\)

\(\Leftrightarrow\frac{1}{2}.2.2019< 4040.\frac{1}{2}\)( nhân hai vế với 1/2)

\(\Leftrightarrow2019< 2020\) ( luôn đúng )

=> điều giả sử đúng

Vậy....

4,Ta có : \(\sqrt{2020}-\sqrt{2019}=\frac{2020-2019}{\sqrt{2020}+\sqrt{2019}}=\frac{1}{\sqrt{2020}+\sqrt{2019}}\)

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

dễ thấy \(2020+2019>2019+2018\Rightarrow\sqrt{2020}+\sqrt{2019}>\sqrt{2019}+\sqrt{2018}\) Từ đó suy ra : \(\frac{1}{\sqrt{2020}+\sqrt{2019}}< \frac{1}{\sqrt{2020}-\sqrt{2019}}\)

theo ss phân số có cùng tử

Vậy....

phần 5 làm tương tự như phần 4 nhé

Ta có: \(\sqrt{2021}-\sqrt{2020}=\frac{\left(\sqrt{2021}-\sqrt{2020}\right)\left(\sqrt{2021}+\sqrt{2020}\right)}{\sqrt{2021}+\sqrt{2020}}=\frac{1}{\sqrt{2021}+\sqrt{2020}}\)

\(\sqrt{2020}-\sqrt{2019}=\frac{\left(\sqrt{2020}+\sqrt{2019}\right)\left(\sqrt{2020}-\sqrt{2019}\right)}{\sqrt{2020}+\sqrt{2019}}=\frac{1}{\sqrt{2020}+\sqrt{2019}}\)

Do \(\frac{1}{\sqrt{2021}+\sqrt{2020}}< \frac{1}{\sqrt{2020}+\sqrt{2019}}\) => \(\sqrt{2021}-\sqrt{2020}< \sqrt{2020}-\sqrt{2019}\)

 \(^6\sqrt{2019} = b, ^6\sqrt{2020} = a \\ Then, A = a^3 - b^3; B = a^2 -b^2\\ \Rightarrow A > B \)

Bài 1: Ta có: \(\sqrt{2020}-\sqrt{2019}=\frac{1}{\sqrt{2020}+\sqrt{2019}};\)\(\sqrt{2018}-\sqrt{2017}=\frac{1}{\sqrt{2018}+\sqrt{2017}}\)

Dễ thấy \(\sqrt{2020}+\sqrt{2019}>\sqrt{2018}+\sqrt{2017}\)nên \(\frac{1}{\sqrt{2020}+\sqrt{2019}}< \frac{1}{\sqrt{2018}+\sqrt{2017}}\)

Suy ra\(\sqrt{2020}-\sqrt{2019}< \sqrt{2018}-\sqrt{2017}\)

Bài 2: Xét biểu thức \(\sqrt{a^2+a^2\left(a+1\right)^2+\left(a+1\right)^2}=\sqrt{a^2\left(a^2+2a+1+1\right)+\left(a+1\right)^2}=\sqrt{a^4+2a^2\left(a+1\right)+\left(a+1\right)^2}=\sqrt{\left(a^2+a+1\right)^2}=a^2+a+1\)(Vì \(a^2+a+1>0\forall a\inℝ\))

Áp dụng công thức tổng quát trên, ta được: \(\sqrt{2019^2+2019^2.2020^2+2020^2}=2019^2+2019+1\)(là số tự nhiên) (đpcm)