Question

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

Answer 1

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

Answer 2

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

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

Có M² = \(\left(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\right)^2\)

           = 8 - 2\(\sqrt{4^2-\left(2\sqrt{3}\right)^2}\)

           = 8 - 2.2 =4

=> M = \(\sqrt{4}\) = 2.

Answer 3

Ta có: \(\dfrac{\sqrt{2}}{\sqrt{2-\sqrt{3}}}-\dfrac{\sqrt{2}}{\sqrt{2+\sqrt{3}}}\)

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

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

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

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

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