Answer
$$\frac{\sqrt{3}+7}{\sqrt{5}+1}=\frac{\sqrt{15}-\sqrt{3}+7\sqrt{5}-7}{4}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{3}+7}{\sqrt{5}+1}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{3}+7}{\sqrt{5}+1}\frac{\sqrt{5}-1}{\sqrt{5}-1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{\sqrt{15}-\sqrt{3}+7\sqrt{5}-7}{5-\sqrt{5}+\sqrt{5}-1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{\sqrt{15}-\sqrt{3}+7\sqrt{5}-7}{4}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{5}-1} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{3} + 7\right) } \cdot \left( \sqrt{5}-1\right) = \color{blue}{ \sqrt{3}} \cdot \sqrt{5}+\color{blue}{ \sqrt{3}} \cdot-1+\color{blue}{7} \cdot \sqrt{5}+\color{blue}{7} \cdot-1 = \\ = \sqrt{15}- \sqrt{3} + 7 \sqrt{5}-7 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{5} + 1\right) } \cdot \left( \sqrt{5}-1\right) = \color{blue}{ \sqrt{5}} \cdot \sqrt{5}+\color{blue}{ \sqrt{5}} \cdot-1+\color{blue}{1} \cdot \sqrt{5}+\color{blue}{1} \cdot-1 = \\ = 5- \sqrt{5} + \sqrt{5}-1 $$ |
| ③ | Simplify numerator and denominator |
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