Answer
$$\frac{3\sqrt{3}}{-\sqrt{3}\cdot9+9\sqrt{2}}=\frac{-3-\sqrt{6}}{3}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{3\sqrt{3}}{-\sqrt{3}\cdot9+9\sqrt{2}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{3\sqrt{3}}{-\sqrt{3}\cdot9+9\sqrt{2}}\frac{-9\sqrt{3}-9\sqrt{2}}{-9\sqrt{3}-9\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{-81-27\sqrt{6}}{243+81\sqrt{6}-81\sqrt{6}-162} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{-81-27\sqrt{6}}{81} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-3-\sqrt{6}}{3}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ - 9 \sqrt{3}- 9 \sqrt{2}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 3 \sqrt{3} } \cdot \left( - 9 \sqrt{3}- 9 \sqrt{2}\right) = \color{blue}{ 3 \sqrt{3}} \cdot- 9 \sqrt{3}+\color{blue}{ 3 \sqrt{3}} \cdot- 9 \sqrt{2} = \\ = -81- 27 \sqrt{6} $$ Simplify denominator. $$ \color{blue}{ \left( - 9 \sqrt{3} + 9 \sqrt{2}\right) } \cdot \left( - 9 \sqrt{3}- 9 \sqrt{2}\right) = \color{blue}{- 9 \sqrt{3}} \cdot- 9 \sqrt{3}\color{blue}{- 9 \sqrt{3}} \cdot- 9 \sqrt{2}+\color{blue}{ 9 \sqrt{2}} \cdot- 9 \sqrt{3}+\color{blue}{ 9 \sqrt{2}} \cdot- 9 \sqrt{2} = \\ = 243 + 81 \sqrt{6}- 81 \sqrt{6}-162 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 27. |
This page was created using
Rationalize Denominator Calculator