Question
Answer
$$\dfrac{7-\sqrt{6}}{6+3\sqrt{6}}=\dfrac{-20+9\sqrt{6}}{6}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{7-\sqrt{6}}{6+3\sqrt{6}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{7-\sqrt{6}}{6+3\sqrt{6}}\frac{6-3\sqrt{6}}{6-3\sqrt{6}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{42-21\sqrt{6}-6\sqrt{6}+18}{36-18\sqrt{6}+18\sqrt{6}-54} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{60-27\sqrt{6}}{-18} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{20-9\sqrt{6}}{-6} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{-20+9\sqrt{6}}{6}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 6- 3 \sqrt{6}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 7- \sqrt{6}\right) } \cdot \left( 6- 3 \sqrt{6}\right) = \color{blue}{7} \cdot6+\color{blue}{7} \cdot- 3 \sqrt{6}\color{blue}{- \sqrt{6}} \cdot6\color{blue}{- \sqrt{6}} \cdot- 3 \sqrt{6} = \\ = 42- 21 \sqrt{6}- 6 \sqrt{6} + 18 $$ Simplify denominator. $$ \color{blue}{ \left( 6 + 3 \sqrt{6}\right) } \cdot \left( 6- 3 \sqrt{6}\right) = \color{blue}{6} \cdot6+\color{blue}{6} \cdot- 3 \sqrt{6}+\color{blue}{ 3 \sqrt{6}} \cdot6+\color{blue}{ 3 \sqrt{6}} \cdot- 3 \sqrt{6} = \\ = 36- 18 \sqrt{6} + 18 \sqrt{6}-54 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 3. |
| ⑤ | Multiply both numerator and denominator by -1. |
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