Answer
$$\frac{\sqrt{3}+1}{\sqrt{5}+2}=\sqrt{15}-2\sqrt{3}+\sqrt{5}-2$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{3}+1}{\sqrt{5}+2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{3}+1}{\sqrt{5}+2}\frac{\sqrt{5}-2}{\sqrt{5}-2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{\sqrt{15}-2\sqrt{3}+\sqrt{5}-2}{5-2\sqrt{5}+2\sqrt{5}-4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{\sqrt{15}-2\sqrt{3}+\sqrt{5}-2}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\sqrt{15}-2\sqrt{3}+\sqrt{5}-2\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{5}-2} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{3} + 1\right) } \cdot \left( \sqrt{5}-2\right) = \color{blue}{ \sqrt{3}} \cdot \sqrt{5}+\color{blue}{ \sqrt{3}} \cdot-2+\color{blue}{1} \cdot \sqrt{5}+\color{blue}{1} \cdot-2 = \\ = \sqrt{15}- 2 \sqrt{3} + \sqrt{5}-2 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{5} + 2\right) } \cdot \left( \sqrt{5}-2\right) = \color{blue}{ \sqrt{5}} \cdot \sqrt{5}+\color{blue}{ \sqrt{5}} \cdot-2+\color{blue}{2} \cdot \sqrt{5}+\color{blue}{2} \cdot-2 = \\ = 5- 2 \sqrt{5} + 2 \sqrt{5}-4 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Remove 1 from denominator. |
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