Question
Answer
$$\dfrac{\sqrt{5}-\sqrt{6}}{\sqrt{3}-\sqrt{10}}=\dfrac{\sqrt{15}-2\sqrt{2}}{7}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{5}-\sqrt{6}}{\sqrt{3}-\sqrt{10}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{5}-\sqrt{6}}{\sqrt{3}-\sqrt{10}}\frac{\sqrt{3}+\sqrt{10}}{\sqrt{3}+\sqrt{10}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{\sqrt{15}+5\sqrt{2}-3\sqrt{2}-2\sqrt{15}}{3+\sqrt{30}-\sqrt{30}-10} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{-\sqrt{15}+2\sqrt{2}}{-7} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{\sqrt{15}-2\sqrt{2}}{7}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{3} + \sqrt{10}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{5}- \sqrt{6}\right) } \cdot \left( \sqrt{3} + \sqrt{10}\right) = \color{blue}{ \sqrt{5}} \cdot \sqrt{3}+\color{blue}{ \sqrt{5}} \cdot \sqrt{10}\color{blue}{- \sqrt{6}} \cdot \sqrt{3}\color{blue}{- \sqrt{6}} \cdot \sqrt{10} = \\ = \sqrt{15} + 5 \sqrt{2}- 3 \sqrt{2}- 2 \sqrt{15} $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{3}- \sqrt{10}\right) } \cdot \left( \sqrt{3} + \sqrt{10}\right) = \color{blue}{ \sqrt{3}} \cdot \sqrt{3}+\color{blue}{ \sqrt{3}} \cdot \sqrt{10}\color{blue}{- \sqrt{10}} \cdot \sqrt{3}\color{blue}{- \sqrt{10}} \cdot \sqrt{10} = \\ = 3 + \sqrt{30}- \sqrt{30}-10 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
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