To calculate: To rationalize the denominator of the expression frac{sqrt{5} - sqrt{2}}{sqrt{5} + sqrt{2}}.

The denominator of a rational expression can be rationalized by removing the irrational term from it.
To remove the irrational term, the numerator and denominator of the expression are multiplied by the conjugate of the denominator.
Conjugate of an expression ${\displaystyle \sqrt{a}+\sqrt{b}}$ is given as ${\displaystyle \sqrt{a}-\sqrt{b}}$ and vice-versa.
Given expression is ${\displaystyle \frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}}$
So teh conjugate of the denominator is given as ${\displaystyle \sqrt{5}-\sqrt{2}}$
Multiplying the numerator and the denominator with the conjugate,
${\displaystyle \frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}=\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}\times \frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}}$
${\displaystyle \frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}=\frac{(\sqrt{5}-\sqrt{2})\times (\sqrt{5}-\sqrt{2})}{(\sqrt{5}+\sqrt{2})\times (\sqrt{5}-\sqrt{2})}}$
${\displaystyle \frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}=\frac{{(\sqrt{5}-\sqrt{2})}^2}{\left({\sqrt{5}}^2\right)-{\left(\sqrt{2}\right)}^2}\text{i. e.}(a+b)(a-b)=a^2-b^2}$
${\displaystyle \frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}=\frac{{\left(\sqrt{5}\right)}^2-2\sqrt{5}\sqrt{2}+{\left(\sqrt{2}\right)}^2}{{\left(5^2\right)}^{{}^1\mathrm{\angle }2}-{\left(2^2\right)}^{{}^1\mathrm{\angle }2}}\text{i. e}{(a-b)}^2=a^2-2ab+b^2}$
$\frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt{2}}=\frac{(\sqrt{5}{)}^2-2\sqrt{5}\sqrt{2}+(\sqrt{2}{)}^2}{(5^2{)}^{{}^1{/}_2}-(2^2{)}^{{}^1{/}_2}}$ According to power rule of exponents $(a^m{)}^n=a^{mn}$

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