Question
Answer
$$\dfrac{578+408\sqrt{2}}{17+12\sqrt{2}}=34$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{578+408\sqrt{2}}{17+12\sqrt{2}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{578+408\sqrt{2}}{17+12\sqrt{2}}\frac{17-12\sqrt{2}}{17-12\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{9826-6936\sqrt{2}+6936\sqrt{2}-9792}{289-204\sqrt{2}+204\sqrt{2}-288} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{34}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}34\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 17- 12 \sqrt{2}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 578 + 408 \sqrt{2}\right) } \cdot \left( 17- 12 \sqrt{2}\right) = \color{blue}{578} \cdot17+\color{blue}{578} \cdot- 12 \sqrt{2}+\color{blue}{ 408 \sqrt{2}} \cdot17+\color{blue}{ 408 \sqrt{2}} \cdot- 12 \sqrt{2} = \\ = 9826- 6936 \sqrt{2} + 6936 \sqrt{2}-9792 $$ Simplify denominator. $$ \color{blue}{ \left( 17 + 12 \sqrt{2}\right) } \cdot \left( 17- 12 \sqrt{2}\right) = \color{blue}{17} \cdot17+\color{blue}{17} \cdot- 12 \sqrt{2}+\color{blue}{ 12 \sqrt{2}} \cdot17+\color{blue}{ 12 \sqrt{2}} \cdot- 12 \sqrt{2} = \\ = 289- 204 \sqrt{2} + 204 \sqrt{2}-288 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Remove 1 from denominator. |
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