Answer
$$\frac{-1}{-2-\sqrt{7}}=\frac{-2+\sqrt{7}}{3}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{-1}{-2-\sqrt{7}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{-1}{-2-\sqrt{7}}\frac{-2+\sqrt{7}}{-2+\sqrt{7}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{2-\sqrt{7}}{4-2\sqrt{7}+2\sqrt{7}-7} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{2-\sqrt{7}}{-3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-2+\sqrt{7}}{3}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ -2 + \sqrt{7}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ -1 } \cdot \left( -2 + \sqrt{7}\right) = \color{blue}{-1} \cdot-2\color{blue}{-1} \cdot \sqrt{7} = \\ = 2- \sqrt{7} $$ Simplify denominator. $$ \color{blue}{ \left( -2- \sqrt{7}\right) } \cdot \left( -2 + \sqrt{7}\right) = \color{blue}{-2} \cdot-2\color{blue}{-2} \cdot \sqrt{7}\color{blue}{- \sqrt{7}} \cdot-2\color{blue}{- \sqrt{7}} \cdot \sqrt{7} = \\ = 4- 2 \sqrt{7} + 2 \sqrt{7}-7 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
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