Giải:

Ta có :

\(Sn=\frac{4n+\sqrt{\left(2n+1\right)\left(2n-1\right)}}{\sqrt{2n+1}+\sqrt{2n-1}}\)

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

\(=\frac{\left(\sqrt{2n+1}\right)^3-\left(\sqrt{2n-1}\right)^3}{2}\)

Tương tự =>\(S_1+S_2+...+S_{40}=\frac{\left(\sqrt{2n_1+1}\right)^3+\sqrt{2n_{40}+1}^3}{2}\)

Sau đó thì dễ rồi ha

Cái đề thấy sai sai. You xem lại thử nhé

\(\sqrt{10+2\sqrt{2}\cdot\sqrt{3}+2\sqrt{2}\cdot\sqrt{5}+2\sqrt{5}\cdot\sqrt{3}}\)=\(\sqrt{\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)^2}=\sqrt{2}+\sqrt{3}+\sqrt{5}\)

ban tu ve hinh nha 

qua A ke duong thang  vuong goc voi MN cat DC tai K (K thuoc DC)

trong tam giác vuông ANK , áp dụng hệ thức lượng ta có

\(\frac{1}{AD^2}=\frac{1}{AN^2}+\frac{1}{AK^2}\)(1)

de dang chung minh duoc tam giac vuong ADK ~tam giac vuong ABM

suy ra \(\frac{AK}{AM}=\frac{AD}{AB}=\frac{1}{2}\) HAY \(AK=\frac{AM}{2}\)

thay vao (1) ta co\(\frac{1}{AD^2}=\frac{1}{AN^2}+\frac{1}{AK^2}\)

\(\frac{1}{\left(\frac{AB}{2}\right)^2}=\frac{1}{AN^2}+\frac{1}{\left(\frac{AM}{2}\right)}^2\)

\(\frac{4}{AB^2}=\frac{1}{AN^2}+\frac{4}{AM^2}\)

\(\sqrt{x^2+x+1}+\sqrt{x^2-x+1}\ge2\sqrt[4]{x^4+x^2+1}\ge2>\frac{1}{2}.\\ \)
\(=>.\) VẬY PHƯƠNG TRÌNH VÔ NGHIỆM.
 

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

\(=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)

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

Nhầm , cho sửa lại đề : 

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