Answer
$$\frac{\sqrt{7}-5}{\sqrt{7}+\sqrt{8}}=-7+2\sqrt{14}+5\sqrt{7}-10\sqrt{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{7}-5}{\sqrt{7}+\sqrt{8}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{7}-5}{\sqrt{7}+\sqrt{8}}\frac{\sqrt{7}-\sqrt{8}}{\sqrt{7}-\sqrt{8}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{7-2\sqrt{14}-5\sqrt{7}+10\sqrt{2}}{7-2\sqrt{14}+2\sqrt{14}-8} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{7-2\sqrt{14}-5\sqrt{7}+10\sqrt{2}}{-1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-7+2\sqrt{14}+5\sqrt{7}-10\sqrt{2}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}-7+2\sqrt{14}+5\sqrt{7}-10\sqrt{2}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{7}- \sqrt{8}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{7}-5\right) } \cdot \left( \sqrt{7}- \sqrt{8}\right) = \color{blue}{ \sqrt{7}} \cdot \sqrt{7}+\color{blue}{ \sqrt{7}} \cdot- \sqrt{8}\color{blue}{-5} \cdot \sqrt{7}\color{blue}{-5} \cdot- \sqrt{8} = \\ = 7- 2 \sqrt{14}- 5 \sqrt{7} + 10 \sqrt{2} $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{7} + \sqrt{8}\right) } \cdot \left( \sqrt{7}- \sqrt{8}\right) = \color{blue}{ \sqrt{7}} \cdot \sqrt{7}+\color{blue}{ \sqrt{7}} \cdot- \sqrt{8}+\color{blue}{ \sqrt{8}} \cdot \sqrt{7}+\color{blue}{ \sqrt{8}} \cdot- \sqrt{8} = \\ = 7- 2 \sqrt{14} + 2 \sqrt{14}-8 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
| ⑤ | Remove 1 from denominator. |
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Simplify Radical Expression