Question
Answer
$$\dfrac{\sqrt{3}}{10+\sqrt{7}}=\dfrac{10\sqrt{3}-\sqrt{21}}{93}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{3}}{10+\sqrt{7}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{3}}{10+\sqrt{7}}\frac{10-\sqrt{7}}{10-\sqrt{7}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{10\sqrt{3}-\sqrt{21}}{100-10\sqrt{7}+10\sqrt{7}-7} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{10\sqrt{3}-\sqrt{21}}{93}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 10- \sqrt{7}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \sqrt{3} } \cdot \left( 10- \sqrt{7}\right) = \color{blue}{ \sqrt{3}} \cdot10+\color{blue}{ \sqrt{3}} \cdot- \sqrt{7} = \\ = 10 \sqrt{3}- \sqrt{21} $$ Simplify denominator. $$ \color{blue}{ \left( 10 + \sqrt{7}\right) } \cdot \left( 10- \sqrt{7}\right) = \color{blue}{10} \cdot10+\color{blue}{10} \cdot- \sqrt{7}+\color{blue}{ \sqrt{7}} \cdot10+\color{blue}{ \sqrt{7}} \cdot- \sqrt{7} = \\ = 100- 10 \sqrt{7} + 10 \sqrt{7}-7 $$ |
| ③ | Simplify numerator and denominator |
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