Answer
$$\frac{1}{\sqrt{13}-\sqrt{12}}=\sqrt{13}+2\sqrt{3}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{1}{\sqrt{13}-\sqrt{12}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1}{\sqrt{13}-\sqrt{12}}\frac{\sqrt{13}+\sqrt{12}}{\sqrt{13}+\sqrt{12}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{\sqrt{13}+2\sqrt{3}}{13+2\sqrt{39}-2\sqrt{39}-12} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{\sqrt{13}+2\sqrt{3}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\sqrt{13}+2\sqrt{3}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{13} + \sqrt{12}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 1 } \cdot \left( \sqrt{13} + \sqrt{12}\right) = \color{blue}{1} \cdot \sqrt{13}+\color{blue}{1} \cdot \sqrt{12} = \\ = \sqrt{13} + 2 \sqrt{3} $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{13}- \sqrt{12}\right) } \cdot \left( \sqrt{13} + \sqrt{12}\right) = \color{blue}{ \sqrt{13}} \cdot \sqrt{13}+\color{blue}{ \sqrt{13}} \cdot \sqrt{12}\color{blue}{- \sqrt{12}} \cdot \sqrt{13}\color{blue}{- \sqrt{12}} \cdot \sqrt{12} = \\ = 13 + 2 \sqrt{39}- 2 \sqrt{39}-12 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Remove 1 from denominator. |
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