Answer
$$\frac{7}{3\sqrt{3}-2\sqrt{2}}=\frac{21\sqrt{3}+14\sqrt{2}}{19}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{7}{3\sqrt{3}-2\sqrt{2}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{7}{3\sqrt{3}-2\sqrt{2}}\frac{3\sqrt{3}+2\sqrt{2}}{3\sqrt{3}+2\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{21\sqrt{3}+14\sqrt{2}}{27+6\sqrt{6}-6\sqrt{6}-8} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{21\sqrt{3}+14\sqrt{2}}{19}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 3 \sqrt{3} + 2 \sqrt{2}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 7 } \cdot \left( 3 \sqrt{3} + 2 \sqrt{2}\right) = \color{blue}{7} \cdot 3 \sqrt{3}+\color{blue}{7} \cdot 2 \sqrt{2} = \\ = 21 \sqrt{3} + 14 \sqrt{2} $$ Simplify denominator. $$ \color{blue}{ \left( 3 \sqrt{3}- 2 \sqrt{2}\right) } \cdot \left( 3 \sqrt{3} + 2 \sqrt{2}\right) = \color{blue}{ 3 \sqrt{3}} \cdot 3 \sqrt{3}+\color{blue}{ 3 \sqrt{3}} \cdot 2 \sqrt{2}\color{blue}{- 2 \sqrt{2}} \cdot 3 \sqrt{3}\color{blue}{- 2 \sqrt{2}} \cdot 2 \sqrt{2} = \\ = 27 + 6 \sqrt{6}- 6 \sqrt{6}-8 $$ |
| ③ | Simplify numerator and denominator |
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