Question
Answer
$$\dfrac{1}{\sqrt{6}-\sqrt{7}}=-\sqrt{6}-\sqrt{7}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{1}{\sqrt{6}-\sqrt{7}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1}{\sqrt{6}-\sqrt{7}}\frac{\sqrt{6}+\sqrt{7}}{\sqrt{6}+\sqrt{7}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{\sqrt{6}+\sqrt{7}}{6+\sqrt{42}-\sqrt{42}-7} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{\sqrt{6}+\sqrt{7}}{-1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}-\frac{\sqrt{6}+\sqrt{7}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ }-(\sqrt{6}+\sqrt{7}) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}-\sqrt{6}-\sqrt{7}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{6} + \sqrt{7}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 1 } \cdot \left( \sqrt{6} + \sqrt{7}\right) = \color{blue}{1} \cdot \sqrt{6}+\color{blue}{1} \cdot \sqrt{7} = \\ = \sqrt{6} + \sqrt{7} $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{6}- \sqrt{7}\right) } \cdot \left( \sqrt{6} + \sqrt{7}\right) = \color{blue}{ \sqrt{6}} \cdot \sqrt{6}+\color{blue}{ \sqrt{6}} \cdot \sqrt{7}\color{blue}{- \sqrt{7}} \cdot \sqrt{6}\color{blue}{- \sqrt{7}} \cdot \sqrt{7} = \\ = 6 + \sqrt{42}- \sqrt{42}-7 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Place a negative sign in front of a fraction. |
| ⑤ | Remove the parenthesis by changing the sign of each term within them. |
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