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