Question
Answer
$$\dfrac{(2+\sqrt{3})^2(2+\sqrt{3})^2+1}{(2+\sqrt{3})^2}=14$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{(2+\sqrt{3})^2(2+\sqrt{3})^2+1}{(2+\sqrt{3})^2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{(4+2\sqrt{3}+2\sqrt{3}+3)(4+2\sqrt{3}+2\sqrt{3}+3)+1}{(2+\sqrt{3})^2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{(7+4\sqrt{3})\cdot(7+4\sqrt{3})+1}{(2+\sqrt{3})^2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{49+28\sqrt{3}+28\sqrt{3}+48+1}{(2+\sqrt{3})^2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}\frac{97+56\sqrt{3}+1}{(2+\sqrt{3})^2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle7}{\textcircled {7}} \htmlClass{explanationCircle explanationCircle8}{\textcircled {8}} } }}}\frac{98+56\sqrt{3}}{4+2\sqrt{3}+2\sqrt{3}+3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle9}{\textcircled {9}} } }}}\frac{98+56\sqrt{3}}{7+4\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle10}{\textcircled {10}} } }}}\frac{98+56\sqrt{3}}{7+4\sqrt{3}}\frac{7-4\sqrt{3}}{7-4\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle11}{\textcircled {11}} } }}}\frac{686-392\sqrt{3}+392\sqrt{3}-672}{49-28\sqrt{3}+28\sqrt{3}-48} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle12}{\textcircled {12}} } }}}\frac{14}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle13}{\textcircled {13}} } }}}14\end{aligned} $$ | |
| ① | $$ (2+\sqrt{3})^2 = \left( 2 + \sqrt{3} \right) \cdot \left( 2 + \sqrt{3} \right) = 4 + 2 \sqrt{3} + 2 \sqrt{3} + 3 $$ |
| ② | $$ (2+\sqrt{3})^2 = \left( 2 + \sqrt{3} \right) \cdot \left( 2 + \sqrt{3} \right) = 4 + 2 \sqrt{3} + 2 \sqrt{3} + 3 $$ |
| ③ | Combine like terms |
| ④ | Combine like terms |
| ⑤ | $$ \color{blue}{ \left( 7 + 4 \sqrt{3}\right) } \cdot \left( 7 + 4 \sqrt{3}\right) = \color{blue}{7} \cdot7+\color{blue}{7} \cdot 4 \sqrt{3}+\color{blue}{ 4 \sqrt{3}} \cdot7+\color{blue}{ 4 \sqrt{3}} \cdot 4 \sqrt{3} = \\ = 49 + 28 \sqrt{3} + 28 \sqrt{3} + 48 $$ |
| ⑥ | Combine like terms |
| ⑦ | Combine like terms |
| ⑧ | $$ (2+\sqrt{3})^2 = \left( 2 + \sqrt{3} \right) \cdot \left( 2 + \sqrt{3} \right) = 4 + 2 \sqrt{3} + 2 \sqrt{3} + 3 $$ |
| ⑨ | Simplify numerator and denominator |
| ⑩ | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 7- 4 \sqrt{3}} $$. |
| ⑪ | Multiply in a numerator. $$ \color{blue}{ \left( 98 + 56 \sqrt{3}\right) } \cdot \left( 7- 4 \sqrt{3}\right) = \color{blue}{98} \cdot7+\color{blue}{98} \cdot- 4 \sqrt{3}+\color{blue}{ 56 \sqrt{3}} \cdot7+\color{blue}{ 56 \sqrt{3}} \cdot- 4 \sqrt{3} = \\ = 686- 392 \sqrt{3} + 392 \sqrt{3}-672 $$ Simplify denominator. $$ \color{blue}{ \left( 7 + 4 \sqrt{3}\right) } \cdot \left( 7- 4 \sqrt{3}\right) = \color{blue}{7} \cdot7+\color{blue}{7} \cdot- 4 \sqrt{3}+\color{blue}{ 4 \sqrt{3}} \cdot7+\color{blue}{ 4 \sqrt{3}} \cdot- 4 \sqrt{3} = \\ = 49- 28 \sqrt{3} + 28 \sqrt{3}-48 $$ |
| ⑫ | Simplify numerator and denominator |
| ⑬ | Remove 1 from denominator. |
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