Question
Answer
$$\dfrac{\sqrt{5}}{\sqrt{15}-\sqrt{10}}=\sqrt{3}+\sqrt{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{5}}{\sqrt{15}-\sqrt{10}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{5}}{\sqrt{15}-\sqrt{10}}\frac{\sqrt{15}+\sqrt{10}}{\sqrt{15}+\sqrt{10}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{5\sqrt{3}+5\sqrt{2}}{15+5\sqrt{6}-5\sqrt{6}-10} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{5\sqrt{3}+5\sqrt{2}}{5} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{\sqrt{3}+\sqrt{2}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\sqrt{3}+\sqrt{2}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{15} + \sqrt{10}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \sqrt{5} } \cdot \left( \sqrt{15} + \sqrt{10}\right) = \color{blue}{ \sqrt{5}} \cdot \sqrt{15}+\color{blue}{ \sqrt{5}} \cdot \sqrt{10} = \\ = 5 \sqrt{3} + 5 \sqrt{2} $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{15}- \sqrt{10}\right) } \cdot \left( \sqrt{15} + \sqrt{10}\right) = \color{blue}{ \sqrt{15}} \cdot \sqrt{15}+\color{blue}{ \sqrt{15}} \cdot \sqrt{10}\color{blue}{- \sqrt{10}} \cdot \sqrt{15}\color{blue}{- \sqrt{10}} \cdot \sqrt{10} = \\ = 15 + 5 \sqrt{6}- 5 \sqrt{6}-10 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 5. |
| ⑤ | Remove 1 from denominator. |
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