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 \(N=2018\frac{1009}{2020}.\)

1. Kẻ \(BH\perp AC\Rightarrow BH=AB.sin60^0=2\sqrt{2}.\frac{\sqrt{3}}{2}=\sqrt{6}\)

\(\Rightarrow S_{ABC}=\frac{1}{2}BH.AC=3\sqrt{2}\)

2. \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(\left(n+1\right)\sqrt{n}+n\sqrt{n+1}\right)\left(\left(n+1\right)\sqrt{n}-n\sqrt{n+1}\right)}\)

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

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

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

1.

S_ABC = \(\frac{AB.AC.sin\widehat{BAC}}{2}\) = ...

\(\forall k\ge0\)ta có :

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

Bạn áp dụng công thức này vào dãy trên ta sẽ có các số hạng triệt tiêu đi nhau và ra kết quả

Câu b dễ hơn nên em xí trước. Nhưng em không chắc đâu:v

b) Xét số hạng tổng quát \(\frac{1}{\sqrt{x}+\sqrt{x+1}}=\frac{\sqrt{x+1}-\sqrt{x}}{\left(\sqrt{x}+\sqrt{x+1}\right)\left(\sqrt{x+1}-\sqrt{x}\right)}=\sqrt{x+1}-\sqrt{x}\) với x >= 0

Áp dụng vào,ta có:

\(A=\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+....+\frac{1}{\sqrt{2019}+\sqrt{2020}}\)

\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{2020}-\sqrt{2019}\)

\(=\sqrt{2020}-1\)

a) \(\frac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{1}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)

\(=\frac{\sqrt{2}}{2+\sqrt{4+2\sqrt{3}}}+\frac{\sqrt{2}}{2-\sqrt{4-2\sqrt{3}}}\)

\(=\frac{\sqrt{2}}{3+\sqrt{3}}+\frac{\sqrt{2}}{3-\sqrt{3}}=\frac{3\sqrt{2}-\sqrt{6}+3\sqrt{2}+\sqrt{6}}{9-3}=\frac{6\sqrt{2}}{6}=\sqrt{2}\)

bn có thể tham khảo ở sách vũ hữu binh nha