Tap the blue circles to see an explanation.
$$ \begin{aligned}\frac{1}{(2+\sqrt{3})^2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1}{4+2\sqrt{3}+2\sqrt{3}+3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{1}{7+4\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{1}{7+4\sqrt{3}}\frac{7-4\sqrt{3}}{7-4\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{7-4\sqrt{3}}{49-28\sqrt{3}+28\sqrt{3}-48} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{7-4\sqrt{3}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}7-4\sqrt{3}\end{aligned} $$ | |
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 | Simplify numerator and denominator |
③ | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 7- 4 \sqrt{3}} $$. |
④ | Multiply in a numerator. $$ \color{blue}{ 1 } \cdot \left( 7- 4 \sqrt{3}\right) = \color{blue}{1} \cdot7+\color{blue}{1} \cdot- 4 \sqrt{3} = \\ = 7- 4 \sqrt{3} $$ Simplify denominator. $$ \color{blue}{ \left( 7 + 4 \sqrt{3}\right) } \cdot \left( 7- 4 \sqrt{3}\right) = \color{blue}{7} \cdot7+\color{blue}{7} \cdot- 4 \sqrt{3}+\color{blue}{ 4 \sqrt{3}} \cdot7+\color{blue}{ 4 \sqrt{3}} \cdot- 4 \sqrt{3} = \\ = 49- 28 \sqrt{3} + 28 \sqrt{3}-48 $$ |
⑤ | Simplify numerator and denominator |
⑥ | Remove 1 from denominator. |