Question
Answer
$$\dfrac{1-\sqrt{2}}{(2\sqrt{2}+1)^2}=\dfrac{17-13\sqrt{2}}{49}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{1-\sqrt{2}}{(2\sqrt{2}+1)^2}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1-\sqrt{2}}{8+2\sqrt{2}+2\sqrt{2}+1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{1-\sqrt{2}}{9+4\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{1-\sqrt{2}}{9+4\sqrt{2}}\frac{9-4\sqrt{2}}{9-4\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{9-4\sqrt{2}-9\sqrt{2}+8}{81-36\sqrt{2}+36\sqrt{2}-32} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{17-13\sqrt{2}}{49}\end{aligned} $$ | |
| ① | $$ (2\sqrt{2}+1)^2 = \left( 2 \sqrt{2} + 1 \right) \cdot \left( 2 \sqrt{2} + 1 \right) = 8 + 2 \sqrt{2} + 2 \sqrt{2} + 1 $$ |
| ② | Simplify numerator and denominator |
| ③ | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 9- 4 \sqrt{2}} $$. |
| ④ | Multiply in a numerator. $$ \color{blue}{ \left( 1- \sqrt{2}\right) } \cdot \left( 9- 4 \sqrt{2}\right) = \color{blue}{1} \cdot9+\color{blue}{1} \cdot- 4 \sqrt{2}\color{blue}{- \sqrt{2}} \cdot9\color{blue}{- \sqrt{2}} \cdot- 4 \sqrt{2} = \\ = 9- 4 \sqrt{2}- 9 \sqrt{2} + 8 $$ Simplify denominator. $$ \color{blue}{ \left( 9 + 4 \sqrt{2}\right) } \cdot \left( 9- 4 \sqrt{2}\right) = \color{blue}{9} \cdot9+\color{blue}{9} \cdot- 4 \sqrt{2}+\color{blue}{ 4 \sqrt{2}} \cdot9+\color{blue}{ 4 \sqrt{2}} \cdot- 4 \sqrt{2} = \\ = 81- 36 \sqrt{2} + 36 \sqrt{2}-32 $$ |
| ⑤ | Simplify numerator and denominator |
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