\(\sqrt{2019^2+2019^2.2020^2+2020^2}=\sqrt{2019^2+\left(2020-1\right)^2.2020^2+2020^2}=\sqrt{2019^2+2020^4-2.2020.2020^2+2020^2+2020^2}=\sqrt{2020^4+2.2020^2-2.\left(2019+1\right).2020^2+2019^2}=\sqrt{2020^4+2.2020^2-2.2019.2020^2-2.2020^2+2019^2}=\sqrt{2020^4-2.2019.2020^2+2019^2}=\sqrt{\left(2020^2-2019\right)^2}=\left|2020^2-2019\right|=2020^2-2019\)

Vì 2020²-2019\(\in N\)

Vậy \(\sqrt{2019^2+2019^2.2020^2+2020^2}\)\(\in N\)

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)

Ta có bài toán tổng quát sau:Chứng minh rằng tổng \(A=\frac{n+1}{n^2+1}+\frac{n+1}{n^2+2}+....+\frac{n+1}{n^2+n}\)(n số hạng và n>1) không phải là số nguyên dương ta có:

\(1=\frac{n+1}{n^2+1}+\frac{n+1}{n^2+2}+...+\frac{n+1}{n^2+3}< \frac{n+1}{n^2+1}+\frac{n+1}{n^2+2}+....+\frac{n+1}{n^2+n}< \frac{n+1}{n^2}+\frac{n+1}{n^2}\)\(+....+\frac{n+1}{n^2}=2\)

Do đó A không phải là số nguyên dương với n=2019 thì ta có bài toán đã cho

1) Ta có: \(2020^2=\left(2019+1\right)^2=2019^2+2.2019+1.\)

\(\Rightarrow1+2019^2=2020^2-2.2019\)

\(\Rightarrow M=\sqrt{1+2019^2+\frac{2019^2}{2020^2}}+\frac{2019}{2020}=\sqrt{2020^2-2.2019+\frac{2019^2}{2020^2}}+\frac{2019}{2020}\)

\(=\sqrt{2020^2-2.2020.\frac{2019}{2020}+\left(\frac{2019}{2020}\right)^2}+\frac{2019}{2020}\)

\(=\sqrt{\left(2020-\frac{2019}{2020}\right)^2}+\frac{2019}{2020}=2020-\frac{2019}{2020}+\frac{2019}{2020}\)

\(=2020\)

Vậy M=2020.

2) Xét  : \(k\in N;k\ge2\)ta có:

\(\left(1+\frac{1}{k-1}-\frac{1}{k}\right)^2=1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}+\frac{2}{k-1}-\frac{2}{\left(k-1\right)k}-\frac{2}{k}\)

                                          \(=1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}+\frac{2}{k-1}-\frac{2}{k-1}+\frac{2}{k}-\frac{2}{k}\)

\(\Rightarrow\left(1+\frac{1}{k-1}-\frac{1}{k}\right)^2=1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}\)

\(\Rightarrow\sqrt{1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}}=1+\frac{1}{k-1}+\frac{1}{k}\)

Cho \(k=3,4,...,2020.\)Ta có:

\(N=\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+...+\sqrt{1+\frac{1}{2019^2}+\frac{1}{2020^2}}\)

\(=\left(1+\frac{1}{2}-\frac{1}{3}\right)+\left(1+\frac{1}{3}-\frac{1}{4}\right)+...+\left(1+\frac{1}{2018}-\frac{1}{2019}\right)+\left(1+\frac{1}{2019}-\frac{1}{2020}\right)\)

\(=2018+\frac{1}{2}-\frac{1}{2020}=2018\frac{1009}{2020}\)

Vậy \(N=2018\frac{1009}{2020}.\)

bn có thể tham khảo ở sách vũ hữu binh nha

Ta có: 

\(a=1-\frac{2019}{2020}+\left(\frac{2019}{2020}\right)^2-\left(\frac{2019}{2020}\right)^3+...+\left(\frac{2019}{2020}\right)^{2020}\)

=> \(\frac{2019}{2020}.a=\frac{2019}{2020}-\left(\frac{2019}{2020}\right)^2+\left(\frac{2019}{2020}\right)^3-...+\left(\frac{2019}{2020}\right)^{2020}-\left(\frac{2019}{2020}\right)^{2021}\)

Lấy

 \(a+\frac{2019}{2020}a=1-\left(\frac{2019}{2020}\right)^{2021}\)

<=> \(a\left(1+\frac{2019}{2020}\right)=\left[1-\left(\frac{2019}{2020}\right)^{2021}\right]\)

<=> \(a.\frac{4039}{2020}=\left[1-\left(\frac{2019}{2020}\right)^{2021}\right]\)

<=> \(a.=\left[1-\left(\frac{2019}{2020}\right)^{2021}\right].\frac{2020}{4039}\)

Vì : \(0< \left(\frac{2019}{2020}\right)^{2021}< 1\)

=> \(0< 1-\left(\frac{2019}{2020}\right)^{2021}< 1\)

và \(0< \frac{2020}{4039}< 1\)

=> \(0< \left[1-\left(\frac{2019}{2020}\right)^{2021}\right].\frac{2020}{4039}< 1\)

=> 0 < a < 1

=> a không phải là một số nguyên.

toan lop may vay ban ?