Answer
$$\frac{\sqrt{5}+\sqrt{6}}{\sqrt{5}-\sqrt{6}}=-11-2\sqrt{30}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{5}+\sqrt{6}}{\sqrt{5}-\sqrt{6}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{5}+\sqrt{6}}{\sqrt{5}-\sqrt{6}}\frac{\sqrt{5}+\sqrt{6}}{\sqrt{5}+\sqrt{6}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{5+\sqrt{30}+\sqrt{30}+6}{5+\sqrt{30}-\sqrt{30}-6} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{11+2\sqrt{30}}{-1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}-\frac{11+2\sqrt{30}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ }-(11+2\sqrt{30}) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}-11-2\sqrt{30}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{5} + \sqrt{6}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{5} + \sqrt{6}\right) } \cdot \left( \sqrt{5} + \sqrt{6}\right) = \color{blue}{ \sqrt{5}} \cdot \sqrt{5}+\color{blue}{ \sqrt{5}} \cdot \sqrt{6}+\color{blue}{ \sqrt{6}} \cdot \sqrt{5}+\color{blue}{ \sqrt{6}} \cdot \sqrt{6} = \\ = 5 + \sqrt{30} + \sqrt{30} + 6 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{5}- \sqrt{6}\right) } \cdot \left( \sqrt{5} + \sqrt{6}\right) = \color{blue}{ \sqrt{5}} \cdot \sqrt{5}+\color{blue}{ \sqrt{5}} \cdot \sqrt{6}\color{blue}{- \sqrt{6}} \cdot \sqrt{5}\color{blue}{- \sqrt{6}} \cdot \sqrt{6} = \\ = 5 + \sqrt{30}- \sqrt{30}-6 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Place a negative sign in front of a fraction. |
| ⑤ | Remove the parenthesis by changing the sign of each term within them. |
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