Answer
$$\frac{1-\sqrt{5}}{2-\sqrt{45}}=\frac{13-\sqrt{5}}{41}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{1-\sqrt{5}}{2-\sqrt{45}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1-\sqrt{5}}{2-\sqrt{45}}\frac{2+\sqrt{45}}{2+\sqrt{45}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{2+3\sqrt{5}-2\sqrt{5}-15}{4+6\sqrt{5}-6\sqrt{5}-45} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{-13+\sqrt{5}}{-41} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{13-\sqrt{5}}{41}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 2 + \sqrt{45}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 1- \sqrt{5}\right) } \cdot \left( 2 + \sqrt{45}\right) = \color{blue}{1} \cdot2+\color{blue}{1} \cdot \sqrt{45}\color{blue}{- \sqrt{5}} \cdot2\color{blue}{- \sqrt{5}} \cdot \sqrt{45} = \\ = 2 + 3 \sqrt{5}- 2 \sqrt{5}-15 $$ Simplify denominator. $$ \color{blue}{ \left( 2- \sqrt{45}\right) } \cdot \left( 2 + \sqrt{45}\right) = \color{blue}{2} \cdot2+\color{blue}{2} \cdot \sqrt{45}\color{blue}{- \sqrt{45}} \cdot2\color{blue}{- \sqrt{45}} \cdot \sqrt{45} = \\ = 4 + 6 \sqrt{5}- 6 \sqrt{5}-45 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
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