Answer
$$\frac{\sqrt{2}+\sqrt{3}}{2\sqrt{3}+3\sqrt{2}}=\frac{\sqrt{6}}{6}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{2}+\sqrt{3}}{2\sqrt{3}+3\sqrt{2}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{2}+\sqrt{3}}{2\sqrt{3}+3\sqrt{2}}\frac{2\sqrt{3}-3\sqrt{2}}{2\sqrt{3}-3\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{2\sqrt{6}-6+6-3\sqrt{6}}{12-6\sqrt{6}+6\sqrt{6}-18} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{-\sqrt{6}}{-6} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{\sqrt{6}}{6}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 2 \sqrt{3}- 3 \sqrt{2}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{2} + \sqrt{3}\right) } \cdot \left( 2 \sqrt{3}- 3 \sqrt{2}\right) = \color{blue}{ \sqrt{2}} \cdot 2 \sqrt{3}+\color{blue}{ \sqrt{2}} \cdot- 3 \sqrt{2}+\color{blue}{ \sqrt{3}} \cdot 2 \sqrt{3}+\color{blue}{ \sqrt{3}} \cdot- 3 \sqrt{2} = \\ = 2 \sqrt{6}-6 + 6- 3 \sqrt{6} $$ Simplify denominator. $$ \color{blue}{ \left( 2 \sqrt{3} + 3 \sqrt{2}\right) } \cdot \left( 2 \sqrt{3}- 3 \sqrt{2}\right) = \color{blue}{ 2 \sqrt{3}} \cdot 2 \sqrt{3}+\color{blue}{ 2 \sqrt{3}} \cdot- 3 \sqrt{2}+\color{blue}{ 3 \sqrt{2}} \cdot 2 \sqrt{3}+\color{blue}{ 3 \sqrt{2}} \cdot- 3 \sqrt{2} = \\ = 12- 6 \sqrt{6} + 6 \sqrt{6}-18 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
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