Answer
$$\frac{1}{\sqrt{5}-\sqrt{2}}=\frac{\sqrt{5}+\sqrt{2}}{3}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{1}{\sqrt{5}-\sqrt{2}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1}{\sqrt{5}-\sqrt{2}}\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}+\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{\sqrt{5}+\sqrt{2}}{5+\sqrt{10}-\sqrt{10}-2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{\sqrt{5}+\sqrt{2}}{3}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{5} + \sqrt{2}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 1 } \cdot \left( \sqrt{5} + \sqrt{2}\right) = \color{blue}{1} \cdot \sqrt{5}+\color{blue}{1} \cdot \sqrt{2} = \\ = \sqrt{5} + \sqrt{2} $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{5}- \sqrt{2}\right) } \cdot \left( \sqrt{5} + \sqrt{2}\right) = \color{blue}{ \sqrt{5}} \cdot \sqrt{5}+\color{blue}{ \sqrt{5}} \cdot \sqrt{2}\color{blue}{- \sqrt{2}} \cdot \sqrt{5}\color{blue}{- \sqrt{2}} \cdot \sqrt{2} = \\ = 5 + \sqrt{10}- \sqrt{10}-2 $$ |
| ③ | Simplify numerator and denominator |
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