Tap the blue circles to see an explanation.
$$ \begin{aligned}\frac{4}{(3-\sqrt{2})^2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{4}{9-3\sqrt{2}-3\sqrt{2}+2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{4}{11-6\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{4}{11-6\sqrt{2}}\frac{11+6\sqrt{2}}{11+6\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{44+24\sqrt{2}}{121+66\sqrt{2}-66\sqrt{2}-72} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{44+24\sqrt{2}}{49}\end{aligned} $$ | |
① | $$ (3-\sqrt{2})^2 = \left( 3- \sqrt{2} \right) \cdot \left( 3- \sqrt{2} \right) = 9- 3 \sqrt{2}- 3 \sqrt{2} + 2 $$ |
② | Simplify numerator and denominator |
③ | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 11 + 6 \sqrt{2}} $$. |
④ | Multiply in a numerator. $$ \color{blue}{ 4 } \cdot \left( 11 + 6 \sqrt{2}\right) = \color{blue}{4} \cdot11+\color{blue}{4} \cdot 6 \sqrt{2} = \\ = 44 + 24 \sqrt{2} $$ Simplify denominator. $$ \color{blue}{ \left( 11- 6 \sqrt{2}\right) } \cdot \left( 11 + 6 \sqrt{2}\right) = \color{blue}{11} \cdot11+\color{blue}{11} \cdot 6 \sqrt{2}\color{blue}{- 6 \sqrt{2}} \cdot11\color{blue}{- 6 \sqrt{2}} \cdot 6 \sqrt{2} = \\ = 121 + 66 \sqrt{2}- 66 \sqrt{2}-72 $$ |
⑤ | Simplify numerator and denominator |