Question
Answer
$$\dfrac{1}{7+2\sqrt{2}}=\dfrac{7-2\sqrt{2}}{41}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{1}{7+2\sqrt{2}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1}{7+2\sqrt{2}}\frac{7-2\sqrt{2}}{7-2\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{7-2\sqrt{2}}{49-14\sqrt{2}+14\sqrt{2}-8} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{7-2\sqrt{2}}{41}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 7- 2 \sqrt{2}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 1 } \cdot \left( 7- 2 \sqrt{2}\right) = \color{blue}{1} \cdot7+\color{blue}{1} \cdot- 2 \sqrt{2} = \\ = 7- 2 \sqrt{2} $$ Simplify denominator. $$ \color{blue}{ \left( 7 + 2 \sqrt{2}\right) } \cdot \left( 7- 2 \sqrt{2}\right) = \color{blue}{7} \cdot7+\color{blue}{7} \cdot- 2 \sqrt{2}+\color{blue}{ 2 \sqrt{2}} \cdot7+\color{blue}{ 2 \sqrt{2}} \cdot- 2 \sqrt{2} = \\ = 49- 14 \sqrt{2} + 14 \sqrt{2}-8 $$ |
| ③ | Simplify numerator and denominator |
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