Answer
$$\frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}}=\frac{5+\sqrt{21}}{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}}\frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}+\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{7+\sqrt{21}+\sqrt{21}+3}{7+\sqrt{21}-\sqrt{21}-3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{10+2\sqrt{21}}{4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{5+\sqrt{21}}{2}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{7} + \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{7} + \sqrt{3}\right) } \cdot \left( \sqrt{7} + \sqrt{3}\right) = \color{blue}{ \sqrt{7}} \cdot \sqrt{7}+\color{blue}{ \sqrt{7}} \cdot \sqrt{3}+\color{blue}{ \sqrt{3}} \cdot \sqrt{7}+\color{blue}{ \sqrt{3}} \cdot \sqrt{3} = \\ = 7 + \sqrt{21} + \sqrt{21} + 3 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{7}- \sqrt{3}\right) } \cdot \left( \sqrt{7} + \sqrt{3}\right) = \color{blue}{ \sqrt{7}} \cdot \sqrt{7}+\color{blue}{ \sqrt{7}} \cdot \sqrt{3}\color{blue}{- \sqrt{3}} \cdot \sqrt{7}\color{blue}{- \sqrt{3}} \cdot \sqrt{3} = \\ = 7 + \sqrt{21}- \sqrt{21}-3 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 2. |
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