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

Trục căn thức ở mẫu 

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

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

= \(=\frac{2+2\sqrt{1-\frac{8}{9}}}{1+\frac{2\sqrt{2}}{3}-1+\frac{2\sqrt{2}}{3}}\)

\(=\frac{2+2\cdot\frac{1}{3}}{\frac{4\sqrt{2}}{3}}=\frac{\frac{8}{3}}{\frac{4\sqrt{2}}{3}}=\frac{8}{3}\cdot\frac{3}{4\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}=vp\)

=> ĐPCM 

Ta có \(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{47}+\sqrt{48}}=\frac{1-\sqrt{2}}{\left(1-\sqrt{2}\right)\left(1+\sqrt{2}\right)}+\frac{\sqrt{2}-\sqrt{3}}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}+\frac{\sqrt{3}-\sqrt{4}}{\left(\sqrt{3}-\sqrt{4}\right)\left(\sqrt{3}+\sqrt{4}\right)}\)

Ta có:

\(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{47}+\sqrt{48}}\)

\(=\frac{\sqrt{1}-\sqrt{2}}{-1}+\frac{\sqrt{2}-\sqrt{3}}{-1}+\frac{\sqrt{3}-\sqrt{4}}{-1}+...+\frac{\sqrt{47}-\sqrt{48}}{-1}\)

\(=\frac{\sqrt{1}-\sqrt{2}+\sqrt{2}-\sqrt{3}+\sqrt{3}-\sqrt{4}+...+\sqrt{47}-\sqrt{48}}{-1}\)

\(=\frac{\sqrt{1}-\sqrt{48}}{-1}\)

\(=4\sqrt{3}-1\approx5,9>3\left(đpcm\right)\)

+ \(\frac{\sqrt{47}-\sqrt{48}}{\left(\sqrt{47}-\sqrt{48}\right)\left(\sqrt{47}+\sqrt{48}\right)}\)

= \(\frac{1-\sqrt{2}+\sqrt{2}-\sqrt{3}+\sqrt{3}-\sqrt{4}+...+\sqrt{47}-\sqrt{48}}{-1}\)

= \(\sqrt{48}-1\left(1\right)\)

Lại có: \(3=4-1=\sqrt{16}-1\left(2\right)\)

Từ (1) và (2) 

=> \(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{47}+\sqrt{48}}>3\)

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

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

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

\(S_{u\left(n\right)}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{25}}=1-\frac{1}{5}< 1\)

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

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

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

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

\(=\dfrac{1}{2}\left(\sqrt{3}+1-\sqrt{3}+1\right)=1\left(\text{đ}pcm\right)\)