Answer
$$\frac{1}{\sqrt{7}-2}=\frac{\sqrt{7}+2}{3}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{1}{\sqrt{7}-2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1}{\sqrt{7}-2}\frac{\sqrt{7}+2}{\sqrt{7}+2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{\sqrt{7}+2}{7+2\sqrt{7}-2\sqrt{7}-4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{\sqrt{7}+2}{3}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{7} + 2} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 1 } \cdot \left( \sqrt{7} + 2\right) = \color{blue}{1} \cdot \sqrt{7}+\color{blue}{1} \cdot2 = \\ = \sqrt{7} + 2 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{7}-2\right) } \cdot \left( \sqrt{7} + 2\right) = \color{blue}{ \sqrt{7}} \cdot \sqrt{7}+\color{blue}{ \sqrt{7}} \cdot2\color{blue}{-2} \cdot \sqrt{7}\color{blue}{-2} \cdot2 = \\ = 7 + 2 \sqrt{7}- 2 \sqrt{7}-4 $$ |
| ③ | Simplify numerator and denominator |
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