Answer
$$\frac{3-\sqrt{7}}{1+2\sqrt{7}}=\frac{-17+7\sqrt{7}}{27}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{3-\sqrt{7}}{1+2\sqrt{7}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{3-\sqrt{7}}{1+2\sqrt{7}}\frac{1-2\sqrt{7}}{1-2\sqrt{7}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{3-6\sqrt{7}-\sqrt{7}+14}{1-2\sqrt{7}+2\sqrt{7}-28} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{17-7\sqrt{7}}{-27} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-17+7\sqrt{7}}{27}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 1- 2 \sqrt{7}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 3- \sqrt{7}\right) } \cdot \left( 1- 2 \sqrt{7}\right) = \color{blue}{3} \cdot1+\color{blue}{3} \cdot- 2 \sqrt{7}\color{blue}{- \sqrt{7}} \cdot1\color{blue}{- \sqrt{7}} \cdot- 2 \sqrt{7} = \\ = 3- 6 \sqrt{7}- \sqrt{7} + 14 $$ Simplify denominator. $$ \color{blue}{ \left( 1 + 2 \sqrt{7}\right) } \cdot \left( 1- 2 \sqrt{7}\right) = \color{blue}{1} \cdot1+\color{blue}{1} \cdot- 2 \sqrt{7}+\color{blue}{ 2 \sqrt{7}} \cdot1+\color{blue}{ 2 \sqrt{7}} \cdot- 2 \sqrt{7} = \\ = 1- 2 \sqrt{7} + 2 \sqrt{7}-28 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
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