Question

(\(\dfrac{1}{\sqrt{\sqrt{5}+2}}\)+\(\sqrt{\sqrt{5}+2}\)).\(\dfrac{1}{\sqrt{\sqrt{5}+1}}\)-\(\sqrt{3-2\sqrt{2}}\)

Answer 1

\(\left(\dfrac{1}{\sqrt{\sqrt{5}+2}}+\sqrt{\sqrt{5}+2}\right)\cdot\dfrac{1}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}\)

\(=\left(\dfrac{1+\sqrt{5}+2}{\sqrt{\sqrt{5}+2}}\right)\cdot\dfrac{1}{\sqrt{\sqrt{5}+1}}-\left(\sqrt{2}-1\right)\)

\(=\dfrac{3+\sqrt{5}}{\sqrt{\left(\sqrt{5}+2\right)\left(\sqrt{5}+1\right)}}-\left(\sqrt{2}-1\right)\)

\(=\dfrac{3+\sqrt{5}}{\sqrt{5+3\sqrt{5}+2}}-\left(\sqrt{2}-1\right)\)

\(=\dfrac{3+\sqrt{5}}{\sqrt{7+3\sqrt{5}}}-\left(\sqrt{2}-1\right)\)

\(=\dfrac{\sqrt{2}\left(3+\sqrt{5}\right)}{\sqrt{14+6\sqrt{5}}}-\left(\sqrt{2}-1\right)=\sqrt{2}-\left(\sqrt{2}-1\right)=1\)