Question
Answer
$$\dfrac{\sqrt{3}}{\sqrt{13}+3}=\dfrac{\sqrt{39}-3\sqrt{3}}{4}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{3}}{\sqrt{13}+3}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{3}}{\sqrt{13}+3}\frac{\sqrt{13}-3}{\sqrt{13}-3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{\sqrt{39}-3\sqrt{3}}{13-3\sqrt{13}+3\sqrt{13}-9} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{\sqrt{39}-3\sqrt{3}}{4}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{13}-3} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \sqrt{3} } \cdot \left( \sqrt{13}-3\right) = \color{blue}{ \sqrt{3}} \cdot \sqrt{13}+\color{blue}{ \sqrt{3}} \cdot-3 = \\ = \sqrt{39}- 3 \sqrt{3} $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{13} + 3\right) } \cdot \left( \sqrt{13}-3\right) = \color{blue}{ \sqrt{13}} \cdot \sqrt{13}+\color{blue}{ \sqrt{13}} \cdot-3+\color{blue}{3} \cdot \sqrt{13}+\color{blue}{3} \cdot-3 = \\ = 13- 3 \sqrt{13} + 3 \sqrt{13}-9 $$ |
| ③ | Simplify numerator and denominator |
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