Question
Answer
$$\dfrac{5}{\sqrt{7}+7}=\dfrac{-5\sqrt{7}+35}{42}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{5}{\sqrt{7}+7}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{5}{\sqrt{7}+7}\frac{\sqrt{7}-7}{\sqrt{7}-7} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{5\sqrt{7}-35}{7-7\sqrt{7}+7\sqrt{7}-49} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{5\sqrt{7}-35}{-42} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-5\sqrt{7}+35}{42}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{7}-7} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 5 } \cdot \left( \sqrt{7}-7\right) = \color{blue}{5} \cdot \sqrt{7}+\color{blue}{5} \cdot-7 = \\ = 5 \sqrt{7}-35 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{7} + 7\right) } \cdot \left( \sqrt{7}-7\right) = \color{blue}{ \sqrt{7}} \cdot \sqrt{7}+\color{blue}{ \sqrt{7}} \cdot-7+\color{blue}{7} \cdot \sqrt{7}+\color{blue}{7} \cdot-7 = \\ = 7- 7 \sqrt{7} + 7 \sqrt{7}-49 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
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