Question
Answer
$$\dfrac{1+\sqrt{2}}{3-2\sqrt{6}}=-\dfrac{3+2\sqrt{6}+3\sqrt{2}+4\sqrt{3}}{15}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{1+\sqrt{2}}{3-2\sqrt{6}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1+\sqrt{2}}{3-2\sqrt{6}}\frac{3+2\sqrt{6}}{3+2\sqrt{6}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{3+2\sqrt{6}+3\sqrt{2}+4\sqrt{3}}{9+6\sqrt{6}-6\sqrt{6}-24} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{3+2\sqrt{6}+3\sqrt{2}+4\sqrt{3}}{-15} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}-\frac{3+2\sqrt{6}+3\sqrt{2}+4\sqrt{3}}{15}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 3 + 2 \sqrt{6}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 1 + \sqrt{2}\right) } \cdot \left( 3 + 2 \sqrt{6}\right) = \color{blue}{1} \cdot3+\color{blue}{1} \cdot 2 \sqrt{6}+\color{blue}{ \sqrt{2}} \cdot3+\color{blue}{ \sqrt{2}} \cdot 2 \sqrt{6} = \\ = 3 + 2 \sqrt{6} + 3 \sqrt{2} + 4 \sqrt{3} $$ Simplify denominator. $$ \color{blue}{ \left( 3- 2 \sqrt{6}\right) } \cdot \left( 3 + 2 \sqrt{6}\right) = \color{blue}{3} \cdot3+\color{blue}{3} \cdot 2 \sqrt{6}\color{blue}{- 2 \sqrt{6}} \cdot3\color{blue}{- 2 \sqrt{6}} \cdot 2 \sqrt{6} = \\ = 9 + 6 \sqrt{6}- 6 \sqrt{6}-24 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Place a negative sign in front of a fraction. |
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