Answer
$$\frac{2+3\sqrt{2}}{3+2\sqrt{3}}=\frac{-6+4\sqrt{3}-9\sqrt{2}+6\sqrt{6}}{3}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2+3\sqrt{2}}{3+2\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2+3\sqrt{2}}{3+2\sqrt{3}}\frac{3-2\sqrt{3}}{3-2\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{6-4\sqrt{3}+9\sqrt{2}-6\sqrt{6}}{9-6\sqrt{3}+6\sqrt{3}-12} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{6-4\sqrt{3}+9\sqrt{2}-6\sqrt{6}}{-3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-6+4\sqrt{3}-9\sqrt{2}+6\sqrt{6}}{3}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 3- 2 \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 2 + 3 \sqrt{2}\right) } \cdot \left( 3- 2 \sqrt{3}\right) = \color{blue}{2} \cdot3+\color{blue}{2} \cdot- 2 \sqrt{3}+\color{blue}{ 3 \sqrt{2}} \cdot3+\color{blue}{ 3 \sqrt{2}} \cdot- 2 \sqrt{3} = \\ = 6- 4 \sqrt{3} + 9 \sqrt{2}- 6 \sqrt{6} $$ Simplify denominator. $$ \color{blue}{ \left( 3 + 2 \sqrt{3}\right) } \cdot \left( 3- 2 \sqrt{3}\right) = \color{blue}{3} \cdot3+\color{blue}{3} \cdot- 2 \sqrt{3}+\color{blue}{ 2 \sqrt{3}} \cdot3+\color{blue}{ 2 \sqrt{3}} \cdot- 2 \sqrt{3} = \\ = 9- 6 \sqrt{3} + 6 \sqrt{3}-12 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
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