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

\(H=\frac{\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)^2}{\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)}\)\(-\frac{\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)^2}{\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)}\)(cái này cùng dòng với cái phía trên)

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

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

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

\(H=\frac{-2}{\sqrt{3}}\)

\(H=-\frac{2\sqrt{3}}{3}\)

Đặt \(A=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)

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

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

\(A^2=4+2=6\)

\(A=\sqrt{6}\)

Đặt \(B=\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\)

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

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

\(B^2=4-2\sqrt{1}=4-2=2\)

\(B=\sqrt{2}\)

Thay vào H 

\(\Rightarrow H=\frac{\sqrt{2}}{\sqrt{6}}-\frac{\sqrt{6}}{\sqrt{2}}=\frac{1}{\sqrt{3}}-\sqrt{3}=\frac{1-3}{\sqrt{3}}=\frac{-2}{\sqrt{3}}\)

học ghê v, đến tận đây r à :V

\(R=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\left(\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\right)\\ =\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{\left(2+\sqrt{2+\sqrt{2+\sqrt{3}}}\right)\left(2-\sqrt{2+\sqrt{2+\sqrt{3}}}\right)}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2^2-\left(\sqrt{2+\sqrt{2+\sqrt{3}}}\right)^2}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{4-\left(2+\sqrt{2+\sqrt{3}}\right)}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2-\sqrt{2+\sqrt{3}}}\)

\(=\sqrt{2+\sqrt{3}}.\left(\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2-\sqrt{2+\sqrt{3}}}\right)\\ =\sqrt{2+\sqrt{3}}.\sqrt{\left(2+\sqrt{2+\sqrt{3}}\right)\left(2-\sqrt{2+\sqrt{3}}\right)}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2^2-\left(\sqrt{2+\sqrt{3}}\right)^2}\\ =\sqrt{2+\sqrt{3}}.\sqrt{4-\left(2+\sqrt{3}\right)}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}\\ =\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}\\ =\sqrt{4-\sqrt{3^2}}\\ =\sqrt{4-3}\\ =\sqrt{1}\\ =1\)

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

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

e có phát hiện mới:v cj chung lớp vs cj kia đúng hemm:v

\(R=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\left(\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\right)\\ =\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{\left(2+\sqrt{2+\sqrt{2+\sqrt{3}}}\right)\left(2-\sqrt{2+\sqrt{2+\sqrt{3}}}\right)}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2^2-\left(\sqrt{2+\sqrt{2+\sqrt{3}}}\right)^2}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{4-\left(2+\sqrt{2+\sqrt{3}}\right)}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2-\sqrt{2+\sqrt{3}}}\)

\(=\sqrt{2+\sqrt{3}}.\left(\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2-\sqrt{2+\sqrt{3}}}\right)\\ =\sqrt{2+\sqrt{3}}.\sqrt{\left(2+\sqrt{2+\sqrt{3}}\right)\left(2-\sqrt{2+\sqrt{3}}\right)}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2^2-\left(\sqrt{2+\sqrt{3}}\right)^2}\\ =\sqrt{2+\sqrt{3}}.\sqrt{4-\left(2+\sqrt{3}\right)}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}\\ =\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}\\ =\sqrt{4-\sqrt{3^2}}\\ =\sqrt{4-3}\\ =\sqrt{1}\\ =1\)

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

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

=1

Lời giải:

Đặt \(\sqrt{2+\sqrt{3}+\sqrt{2-\sqrt{3}}}=a; \sqrt{2+\sqrt{3}-\sqrt{2-\sqrt{3}}}=b\)

Có:

\(a^2+b^2=(2+\sqrt{3}+\sqrt{2-\sqrt{3}})+(2+\sqrt{3}-\sqrt{2-\sqrt{3}})=2(2+\sqrt{3})\)

\(=4+2\sqrt{3}=3+1+2\sqrt{3.1}=(\sqrt{3}+1)^2\)

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

\(=\sqrt{(2+\sqrt{3})^2-(2-\sqrt{3})}=\sqrt{5+5\sqrt{3}}\)

Như vậy:

\(\frac{\sqrt{2+\sqrt{3}+\sqrt{2-\sqrt{3}}}}{\sqrt{2+\sqrt{3}-\sqrt{2-\sqrt{3}}}}+\frac{\sqrt{2+\sqrt{3}-\sqrt{2-\sqrt{3}}}}{\sqrt{2+\sqrt{3}+\sqrt{2-\sqrt{3}}}}=\frac{a}{b}+\frac{b}{a}=\frac{a^2+b^2}{ab}\)

\(=\frac{(\sqrt{3}+1)^2}{\sqrt{5+5\sqrt{3}}}=\frac{(\sqrt{3}+1)^2}{\sqrt{5}.\sqrt{\sqrt{3}+1}}=\frac{(\sqrt{3}+1)^{1.5}}{\sqrt{5}}\)

\(\sqrt{2+\sqrt{3}.}\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{\left(2+\sqrt{2+\sqrt{2+\sqrt{3}}}\right)\left(2-\sqrt{2+\sqrt{2+\sqrt{3}}}\right)}\)

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

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

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

=\(\sqrt{4-3}=1\)

\(A=\frac{\sqrt{2}\left(\sqrt{3}-3\right)}{\sqrt{4-2\sqrt{3}}+4}+\frac{\sqrt{2}\left(\sqrt{3}+3\right)}{\sqrt{4+2\sqrt{3}}-4}=\frac{\sqrt{2}\left(\sqrt{3}-3\right)}{\sqrt{\left(\sqrt{3}-1\right)^2}+4}+\frac{\sqrt{2}\left(\sqrt{3}+3\right)}{\sqrt{\left(\sqrt{3}+1\right)^2}-4}\)

\(A=\frac{\sqrt{2}\left(\sqrt{3}-3\right)}{3+\sqrt{3}}+\frac{\sqrt{2}\left(\sqrt{3}+3\right)}{\sqrt{3}-3}=\frac{\sqrt{2}\left(\sqrt{3}-3\right)^2+\sqrt{2}\left(\sqrt{3}+3\right)^2}{3-9}=\frac{24\sqrt{2}}{-6}=-4\sqrt{2}\)