Question

Chứng minh:

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

Answer 1

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

= \(\dfrac{2\sqrt{2}+\sqrt{6}}{2+\sqrt{4+2\sqrt{3}}}+\dfrac{2\sqrt{2}-\sqrt{6}}{2-\sqrt{4-2\sqrt{3}}}\)

=\(\dfrac{2\sqrt{2}+\sqrt{6}}{2+1+\sqrt{3}}+\dfrac{2\sqrt{2}-\sqrt{6}}{2+1-\sqrt{3}}\)

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

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

=\(\dfrac{6\sqrt{2}}{6}\)

=\(\sqrt{2}\)