Answer
$$\frac{33}{4-\sqrt{5}}=\frac{132+33\sqrt{5}}{11}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{33}{4-\sqrt{5}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{33}{4-\sqrt{5}}\frac{4+\sqrt{5}}{4+\sqrt{5}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{132+33\sqrt{5}}{16+4\sqrt{5}-4\sqrt{5}-5} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{132+33\sqrt{5}}{11}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 4 + \sqrt{5}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 33 } \cdot \left( 4 + \sqrt{5}\right) = \color{blue}{33} \cdot4+\color{blue}{33} \cdot \sqrt{5} = \\ = 132 + 33 \sqrt{5} $$ Simplify denominator. $$ \color{blue}{ \left( 4- \sqrt{5}\right) } \cdot \left( 4 + \sqrt{5}\right) = \color{blue}{4} \cdot4+\color{blue}{4} \cdot \sqrt{5}\color{blue}{- \sqrt{5}} \cdot4\color{blue}{- \sqrt{5}} \cdot \sqrt{5} = \\ = 16 + 4 \sqrt{5}- 4 \sqrt{5}-5 $$ |
| ③ | Simplify numerator and denominator |
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