Answer
$$\frac{\sqrt{20}+3}{\sqrt{5}+2}=4-\sqrt{5}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{20}+3}{\sqrt{5}+2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{20}+3}{\sqrt{5}+2}\frac{\sqrt{5}-2}{\sqrt{5}-2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{10-4\sqrt{5}+3\sqrt{5}-6}{5-2\sqrt{5}+2\sqrt{5}-4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{4-\sqrt{5}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}4-\sqrt{5}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{5}-2} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{20} + 3\right) } \cdot \left( \sqrt{5}-2\right) = \color{blue}{ \sqrt{20}} \cdot \sqrt{5}+\color{blue}{ \sqrt{20}} \cdot-2+\color{blue}{3} \cdot \sqrt{5}+\color{blue}{3} \cdot-2 = \\ = 10- 4 \sqrt{5} + 3 \sqrt{5}-6 $$ 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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