Question
Answer
$$\dfrac{1}{10+\sqrt{3}}=\dfrac{10-\sqrt{3}}{97}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{1}{10+\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1}{10+\sqrt{3}}\frac{10-\sqrt{3}}{10-\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{10-\sqrt{3}}{100-10\sqrt{3}+10\sqrt{3}-3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{10-\sqrt{3}}{97}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 10- \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 1 } \cdot \left( 10- \sqrt{3}\right) = \color{blue}{1} \cdot10+\color{blue}{1} \cdot- \sqrt{3} = \\ = 10- \sqrt{3} $$ Simplify denominator. $$ \color{blue}{ \left( 10 + \sqrt{3}\right) } \cdot \left( 10- \sqrt{3}\right) = \color{blue}{10} \cdot10+\color{blue}{10} \cdot- \sqrt{3}+\color{blue}{ \sqrt{3}} \cdot10+\color{blue}{ \sqrt{3}} \cdot- \sqrt{3} = \\ = 100- 10 \sqrt{3} + 10 \sqrt{3}-3 $$ |
| ③ | Simplify numerator and denominator |
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