Question
Answer
$$\dfrac{\sqrt{300}-\sqrt{3}}{\sqrt{3}}=9$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{300}-\sqrt{3}}{\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{300}-\sqrt{3}}{\sqrt{3}}\frac{\sqrt{3}}{\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{30-3}{3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{27}{3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}} \frac{ 27 : \color{orangered}{ 3 } }{ 3 : \color{orangered}{ 3 }} \xlongequal{ } \\[1 em] & \xlongequal{ }\frac{9}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}9\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{300}- \sqrt{3}\right) } \cdot \sqrt{3} = \color{blue}{ \sqrt{300}} \cdot \sqrt{3}\color{blue}{- \sqrt{3}} \cdot \sqrt{3} = \\ = 30-3 $$ Simplify denominator. $$ \color{blue}{ \sqrt{3} } \cdot \sqrt{3} = 3 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both the top and bottom numbers by $ \color{orangered}{ 3 } $. |
| ⑤ | Remove 1 from denominator. |
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