Answer
$$\frac{9}{2\sqrt{7}+1}=\frac{2\sqrt{7}-1}{3}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{9}{2\sqrt{7}+1}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{9}{2\sqrt{7}+1}\frac{2\sqrt{7}-1}{2\sqrt{7}-1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{18\sqrt{7}-9}{28-2\sqrt{7}+2\sqrt{7}-1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{18\sqrt{7}-9}{27} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{2\sqrt{7}-1}{3}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 2 \sqrt{7}-1} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 9 } \cdot \left( 2 \sqrt{7}-1\right) = \color{blue}{9} \cdot 2 \sqrt{7}+\color{blue}{9} \cdot-1 = \\ = 18 \sqrt{7}-9 $$ Simplify denominator. $$ \color{blue}{ \left( 2 \sqrt{7} + 1\right) } \cdot \left( 2 \sqrt{7}-1\right) = \color{blue}{ 2 \sqrt{7}} \cdot 2 \sqrt{7}+\color{blue}{ 2 \sqrt{7}} \cdot-1+\color{blue}{1} \cdot 2 \sqrt{7}+\color{blue}{1} \cdot-1 = \\ = 28- 2 \sqrt{7} + 2 \sqrt{7}-1 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 9. |
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