Question
Answer
$$\dfrac{\sqrt{6}}{\sqrt{17}+3}=\dfrac{\sqrt{102}-3\sqrt{6}}{8}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{6}}{\sqrt{17}+3}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{6}}{\sqrt{17}+3}\frac{\sqrt{17}-3}{\sqrt{17}-3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{\sqrt{102}-3\sqrt{6}}{17-3\sqrt{17}+3\sqrt{17}-9} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{\sqrt{102}-3\sqrt{6}}{8}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{17}-3} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \sqrt{6} } \cdot \left( \sqrt{17}-3\right) = \color{blue}{ \sqrt{6}} \cdot \sqrt{17}+\color{blue}{ \sqrt{6}} \cdot-3 = \\ = \sqrt{102}- 3 \sqrt{6} $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{17} + 3\right) } \cdot \left( \sqrt{17}-3\right) = \color{blue}{ \sqrt{17}} \cdot \sqrt{17}+\color{blue}{ \sqrt{17}} \cdot-3+\color{blue}{3} \cdot \sqrt{17}+\color{blue}{3} \cdot-3 = \\ = 17- 3 \sqrt{17} + 3 \sqrt{17}-9 $$ |
| ③ | Simplify numerator and denominator |
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