Answer
$$\frac{1-\sqrt{3}\cdot(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}=\frac{-2-\sqrt{3}}{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{1-\sqrt{3}\cdot(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{\frac{1-3-\sqrt{3}}{1}}{3+\sqrt{3}-\sqrt{3}-1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-2-\sqrt{3}}{2}\end{aligned} $$ | |
| ① | $$ 1-\sqrt{3}\cdot(\sqrt{3}+1)
= 1 \cdot \color{blue}{\frac{ 1 }{ 1}} - 3+\sqrt{3} \cdot \color{blue}{\frac{ 1 }{ 1}}
= \frac{1-3-\sqrt{3}}{1} $$ |
| ② | $$ \color{blue}{ \left( \sqrt{3}-1\right) } \cdot \left( \sqrt{3} + 1\right) = \color{blue}{ \sqrt{3}} \cdot \sqrt{3}+\color{blue}{ \sqrt{3}} \cdot1\color{blue}{-1} \cdot \sqrt{3}\color{blue}{-1} \cdot1 = \\ = 3 + \sqrt{3}- \sqrt{3}-1 $$ |
| ③ | Remove 1 from denominator. |
| ④ | Combine like terms |
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