Answer
$$\frac{9}{\sqrt{10}+3}=9\sqrt{10}-27$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{9}{\sqrt{10}+3}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{9}{\sqrt{10}+3}\frac{\sqrt{10}-3}{\sqrt{10}-3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{9\sqrt{10}-27}{10-3\sqrt{10}+3\sqrt{10}-9} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{9\sqrt{10}-27}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}9\sqrt{10}-27\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{10}-3} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 9 } \cdot \left( \sqrt{10}-3\right) = \color{blue}{9} \cdot \sqrt{10}+\color{blue}{9} \cdot-3 = \\ = 9 \sqrt{10}-27 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{10} + 3\right) } \cdot \left( \sqrt{10}-3\right) = \color{blue}{ \sqrt{10}} \cdot \sqrt{10}+\color{blue}{ \sqrt{10}} \cdot-3+\color{blue}{3} \cdot \sqrt{10}+\color{blue}{3} \cdot-3 = \\ = 10- 3 \sqrt{10} + 3 \sqrt{10}-9 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Remove 1 from denominator. |
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