Question
Answer
$$\dfrac{\sqrt{5}+3\sqrt{6}}{2+3\sqrt{5}}=\dfrac{-2\sqrt{5}+15-6\sqrt{6}+9\sqrt{30}}{41}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{5}+3\sqrt{6}}{2+3\sqrt{5}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{5}+3\sqrt{6}}{2+3\sqrt{5}}\frac{2-3\sqrt{5}}{2-3\sqrt{5}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{2\sqrt{5}-15+6\sqrt{6}-9\sqrt{30}}{4-6\sqrt{5}+6\sqrt{5}-45} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{2\sqrt{5}-15+6\sqrt{6}-9\sqrt{30}}{-41} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-2\sqrt{5}+15-6\sqrt{6}+9\sqrt{30}}{41}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 2- 3 \sqrt{5}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{5} + 3 \sqrt{6}\right) } \cdot \left( 2- 3 \sqrt{5}\right) = \color{blue}{ \sqrt{5}} \cdot2+\color{blue}{ \sqrt{5}} \cdot- 3 \sqrt{5}+\color{blue}{ 3 \sqrt{6}} \cdot2+\color{blue}{ 3 \sqrt{6}} \cdot- 3 \sqrt{5} = \\ = 2 \sqrt{5}-15 + 6 \sqrt{6}- 9 \sqrt{30} $$ Simplify denominator. $$ \color{blue}{ \left( 2 + 3 \sqrt{5}\right) } \cdot \left( 2- 3 \sqrt{5}\right) = \color{blue}{2} \cdot2+\color{blue}{2} \cdot- 3 \sqrt{5}+\color{blue}{ 3 \sqrt{5}} \cdot2+\color{blue}{ 3 \sqrt{5}} \cdot- 3 \sqrt{5} = \\ = 4- 6 \sqrt{5} + 6 \sqrt{5}-45 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
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