Question

Simplify sqrt((sqrt(5) - sqrt(2))/(sqrt(5) + sqrt(2))) + sqrt((sqrt(5) + sqrt(2))/(sqrt(5) - sqrt(2)))

Answer 1

Let's denote:
\[
x = \sqrt{\frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} + \sqrt{2}}} + \sqrt{\frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}}}
\]

Now, we can simplify each term separately. We start with \( a = \sqrt{\frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} + \sqrt{2}}} \) and \( b = \sqrt{\frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}}} \). Notice that these terms are reciprocals of each other.

Let's calculate \( a^2 \) and \( b^2 \):
\[
a^2 = \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} + \sqrt{2}}, \quad b^2 = \frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}}
\]

Next, we consider \( a^2 + b^2 \):
\[
a^2 + b^2 = \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} + \sqrt{2}} + \frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}}
\]

Now, we find a common denominator for these fractions:
\[
a^2 + b^2 = \frac{(\sqrt{5} - \sqrt{2})^2 + (\sqrt{5} + \sqrt{2})^2}{(\sqrt{5} + \sqrt{2})(\sqrt{5} - \sqrt{2})}
\]

Calculating \((\sqrt{5} - \sqrt{2})^2\) and \((\sqrt{5} + \sqrt{2})^2\):
\[
(\sqrt{5} - \sqrt{2})^2 = 5 - 2\sqrt{10} + 2 = 7 - 2\sqrt{10}
\]
\[
(\sqrt{5} + \sqrt{2})^2 = 5 + 2\sqrt{10} + 2 = 7 + 2\sqrt{10}
\]

Thus,
\[
a^2 + b^2 = \frac{(7 - 2\sqrt{10}) + (7 + 2\sqrt{10})}{(\sqrt{5})^2 - (\sqrt{2})^2} = \frac{14}{5 - 2} = \frac{14}{3}
\]

Also, we can find \( ab \):
\[
ab = \sqrt{a^2} \sqrt{b^2} = \sqrt{\left( \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} + \sqrt{2}} \cdot \frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}} \right)} = \sqrt{1} = 1
\]

Now we can use the identity \( (a + b)^2 = a^2 + b^2 + 2ab \):
\[
x^2 = a^2 + b^2 + 2ab = \frac{14}{3} + 2 \cdot 1 = \frac{14}{3} + 2 = \frac{14}{3} + \frac{6}{3} = \frac{20}{3}
\]

Therefore:
\[
x = \sqrt{\frac{20}{3}} = \frac{\sqrt{20}}{\sqrt{3}} = \frac{2\sqrt{5}}{\sqrt{3}} = \frac{2\sqrt{15}}{3}
\]

Thus, the simplified expression is:
\[
\boxed{\frac{2\sqrt{15}}{3}}
\]