[tex]\dfrac{2}{\sqrt{2+\sqrt2}}=\dfrac{2}{\sqrt{2+\sqrt2}}\cdot\dfrac{\sqrt{2-\sqrt2}}{\sqrt{2-\sqrt2}}[/tex]
use √(ab) = √a · √b
[tex]=\dfrac{2\sqrt{2-\sqrt2}}{\sqrt{(2+\sqrt2)(2-\sqrt2)}}[/tex]
use (a + b)(a - b) = a² - b²
[tex]=\dfrac{2\sqrt{2-\sqrt2}}{\sqrt{2^2-(\sqrt2)^2}}=\dfrac{2\sqrt{2-\sqrt2}}{\sqrt{4-2}}=\dfrac{2\sqrt{2-\sqrt2}}{\sqrt2}\cdot\dfrac{\sqrt2}{\sqrt2}[/tex]
use √a · √a = a and √(ab) = √a · √b
[tex]=\dfrac{2\sqrt{2\cdot2-\sqrt2\cdot2}}{2}=\sqrt{4-2\sqrt2}\\\\Answer:\ \boxed{\dfrac{2}{\sqrt{2+\sqrt2}}=\sqrt{4-2\sqrt2}}[/tex]
