Answer
$$\frac{3}{\sqrt{21}-3}=\frac{\sqrt{21}+3}{4}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{3}{\sqrt{21}-3}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{3}{\sqrt{21}-3}\frac{\sqrt{21}+3}{\sqrt{21}+3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{3\sqrt{21}+9}{21+3\sqrt{21}-3\sqrt{21}-9} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{3\sqrt{21}+9}{12} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{\sqrt{21}+3}{4}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{21} + 3} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 3 } \cdot \left( \sqrt{21} + 3\right) = \color{blue}{3} \cdot \sqrt{21}+\color{blue}{3} \cdot3 = \\ = 3 \sqrt{21} + 9 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{21}-3\right) } \cdot \left( \sqrt{21} + 3\right) = \color{blue}{ \sqrt{21}} \cdot \sqrt{21}+\color{blue}{ \sqrt{21}} \cdot3\color{blue}{-3} \cdot \sqrt{21}\color{blue}{-3} \cdot3 = \\ = 21 + 3 \sqrt{21}- 3 \sqrt{21}-9 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 3. |
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