Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{(\sqrt{3}+4)^2}{3+\sqrt{2}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{3+4\sqrt{3}+4\sqrt{3}+16}{3+\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{19+8\sqrt{3}}{3+\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{19+8\sqrt{3}}{3+\sqrt{2}}\frac{3-\sqrt{2}}{3-\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{57-19\sqrt{2}+24\sqrt{3}-8\sqrt{6}}{9-3\sqrt{2}+3\sqrt{2}-2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{57-19\sqrt{2}+24\sqrt{3}-8\sqrt{6}}{7}\end{aligned} $$ | |
| ① | $$ (\sqrt{3}+4)^2 = \left( \sqrt{3} + 4 \right) \cdot \left( \sqrt{3} + 4 \right) = 3 + 4 \sqrt{3} + 4 \sqrt{3} + 16 $$ |
| ② | Simplify numerator and denominator |
| ③ | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 3- \sqrt{2}} $$. |
| ④ | Multiply in a numerator. $$ \color{blue}{ \left( 19 + 8 \sqrt{3}\right) } \cdot \left( 3- \sqrt{2}\right) = \color{blue}{19} \cdot3+\color{blue}{19} \cdot- \sqrt{2}+\color{blue}{ 8 \sqrt{3}} \cdot3+\color{blue}{ 8 \sqrt{3}} \cdot- \sqrt{2} = \\ = 57- 19 \sqrt{2} + 24 \sqrt{3}- 8 \sqrt{6} $$ Simplify denominator. $$ \color{blue}{ \left( 3 + \sqrt{2}\right) } \cdot \left( 3- \sqrt{2}\right) = \color{blue}{3} \cdot3+\color{blue}{3} \cdot- \sqrt{2}+\color{blue}{ \sqrt{2}} \cdot3+\color{blue}{ \sqrt{2}} \cdot- \sqrt{2} = \\ = 9- 3 \sqrt{2} + 3 \sqrt{2}-2 $$ |
| ⑤ | Simplify numerator and denominator |