Tap the blue circles to see an explanation.
$$ \begin{aligned}\frac{\sqrt{6}+\sqrt{24}}{\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{6}+\sqrt{24}}{\sqrt{3}}\frac{\sqrt{3}}{\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{3\sqrt{2}+6\sqrt{2}}{3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{9\sqrt{2}}{3} \xlongequal{ } \\[1 em] & \xlongequal{ }3\sqrt{2}\end{aligned} $$ | |
① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{3}} $$. |
② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{6} + \sqrt{24}\right) } \cdot \sqrt{3} = \color{blue}{ \sqrt{6}} \cdot \sqrt{3}+\color{blue}{ \sqrt{24}} \cdot \sqrt{3} = \\ = 3 \sqrt{2} + 6 \sqrt{2} $$ Simplify denominator. $$ \color{blue}{ \sqrt{3} } \cdot \sqrt{3} = 3 $$ |
③ | Simplify numerator and denominator |