Question
Answer
$$\dfrac{2\sqrt{6}}{9\sqrt{2}+\sqrt{3}}=\dfrac{36\sqrt{3}-6\sqrt{2}}{159}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2\sqrt{6}}{9\sqrt{2}+\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2\sqrt{6}}{9\sqrt{2}+\sqrt{3}}\frac{9\sqrt{2}-\sqrt{3}}{9\sqrt{2}-\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{36\sqrt{3}-6\sqrt{2}}{162-9\sqrt{6}+9\sqrt{6}-3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{36\sqrt{3}-6\sqrt{2}}{159}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 9 \sqrt{2}- \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 2 \sqrt{6} } \cdot \left( 9 \sqrt{2}- \sqrt{3}\right) = \color{blue}{ 2 \sqrt{6}} \cdot 9 \sqrt{2}+\color{blue}{ 2 \sqrt{6}} \cdot- \sqrt{3} = \\ = 36 \sqrt{3}- 6 \sqrt{2} $$ Simplify denominator. $$ \color{blue}{ \left( 9 \sqrt{2} + \sqrt{3}\right) } \cdot \left( 9 \sqrt{2}- \sqrt{3}\right) = \color{blue}{ 9 \sqrt{2}} \cdot 9 \sqrt{2}+\color{blue}{ 9 \sqrt{2}} \cdot- \sqrt{3}+\color{blue}{ \sqrt{3}} \cdot 9 \sqrt{2}+\color{blue}{ \sqrt{3}} \cdot- \sqrt{3} = \\ = 162- 9 \sqrt{6} + 9 \sqrt{6}-3 $$ |
| ③ | Simplify numerator and denominator |
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