Answer
$$\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}=5-2\sqrt{6}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}-2\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{18-6\sqrt{6}-6\sqrt{6}+12}{18-6\sqrt{6}+6\sqrt{6}-12} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{30-12\sqrt{6}}{6} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{5-2\sqrt{6}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}5-2\sqrt{6}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 3 \sqrt{2}- 2 \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 3 \sqrt{2}- 2 \sqrt{3}\right) } \cdot \left( 3 \sqrt{2}- 2 \sqrt{3}\right) = \color{blue}{ 3 \sqrt{2}} \cdot 3 \sqrt{2}+\color{blue}{ 3 \sqrt{2}} \cdot- 2 \sqrt{3}\color{blue}{- 2 \sqrt{3}} \cdot 3 \sqrt{2}\color{blue}{- 2 \sqrt{3}} \cdot- 2 \sqrt{3} = \\ = 18- 6 \sqrt{6}- 6 \sqrt{6} + 12 $$ Simplify denominator. $$ \color{blue}{ \left( 3 \sqrt{2} + 2 \sqrt{3}\right) } \cdot \left( 3 \sqrt{2}- 2 \sqrt{3}\right) = \color{blue}{ 3 \sqrt{2}} \cdot 3 \sqrt{2}+\color{blue}{ 3 \sqrt{2}} \cdot- 2 \sqrt{3}+\color{blue}{ 2 \sqrt{3}} \cdot 3 \sqrt{2}+\color{blue}{ 2 \sqrt{3}} \cdot- 2 \sqrt{3} = \\ = 18- 6 \sqrt{6} + 6 \sqrt{6}-12 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 6. |
| ⑤ | Remove 1 from denominator. |
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