Question
Answer
$$\dfrac{3+\sqrt{7}}{2+2\sqrt{7}}=\dfrac{2+\sqrt{7}}{6}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{3+\sqrt{7}}{2+2\sqrt{7}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{3+\sqrt{7}}{2+2\sqrt{7}}\frac{2-2\sqrt{7}}{2-2\sqrt{7}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{6-6\sqrt{7}+2\sqrt{7}-14}{4-4\sqrt{7}+4\sqrt{7}-28} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{-8-4\sqrt{7}}{-24} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-2-\sqrt{7}}{-6} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{2+\sqrt{7}}{6}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 2- 2 \sqrt{7}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 3 + \sqrt{7}\right) } \cdot \left( 2- 2 \sqrt{7}\right) = \color{blue}{3} \cdot2+\color{blue}{3} \cdot- 2 \sqrt{7}+\color{blue}{ \sqrt{7}} \cdot2+\color{blue}{ \sqrt{7}} \cdot- 2 \sqrt{7} = \\ = 6- 6 \sqrt{7} + 2 \sqrt{7}-14 $$ Simplify denominator. $$ \color{blue}{ \left( 2 + 2 \sqrt{7}\right) } \cdot \left( 2- 2 \sqrt{7}\right) = \color{blue}{2} \cdot2+\color{blue}{2} \cdot- 2 \sqrt{7}+\color{blue}{ 2 \sqrt{7}} \cdot2+\color{blue}{ 2 \sqrt{7}} \cdot- 2 \sqrt{7} = \\ = 4- 4 \sqrt{7} + 4 \sqrt{7}-28 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 4. |
| ⑤ | Multiply both numerator and denominator by -1. |
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