Question
Answer
$$\dfrac{\sqrt{3}}{6+\sqrt{2}}=\dfrac{6\sqrt{3}-\sqrt{6}}{34}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{3}}{6+\sqrt{2}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{3}}{6+\sqrt{2}}\frac{6-\sqrt{2}}{6-\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{6\sqrt{3}-\sqrt{6}}{36-6\sqrt{2}+6\sqrt{2}-2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{6\sqrt{3}-\sqrt{6}}{34}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 6- \sqrt{2}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \sqrt{3} } \cdot \left( 6- \sqrt{2}\right) = \color{blue}{ \sqrt{3}} \cdot6+\color{blue}{ \sqrt{3}} \cdot- \sqrt{2} = \\ = 6 \sqrt{3}- \sqrt{6} $$ Simplify denominator. $$ \color{blue}{ \left( 6 + \sqrt{2}\right) } \cdot \left( 6- \sqrt{2}\right) = \color{blue}{6} \cdot6+\color{blue}{6} \cdot- \sqrt{2}+\color{blue}{ \sqrt{2}} \cdot6+\color{blue}{ \sqrt{2}} \cdot- \sqrt{2} = \\ = 36- 6 \sqrt{2} + 6 \sqrt{2}-2 $$ |
| ③ | Simplify numerator and denominator |
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