Question
Answer
$$\dfrac{3+\sqrt{2}}{5+\sqrt{8}}=\dfrac{11-\sqrt{2}}{17}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{3+\sqrt{2}}{5+\sqrt{8}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{3+\sqrt{2}}{5+\sqrt{8}}\frac{5-\sqrt{8}}{5-\sqrt{8}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{15-6\sqrt{2}+5\sqrt{2}-4}{25-10\sqrt{2}+10\sqrt{2}-8} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{11-\sqrt{2}}{17}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 5- \sqrt{8}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 3 + \sqrt{2}\right) } \cdot \left( 5- \sqrt{8}\right) = \color{blue}{3} \cdot5+\color{blue}{3} \cdot- \sqrt{8}+\color{blue}{ \sqrt{2}} \cdot5+\color{blue}{ \sqrt{2}} \cdot- \sqrt{8} = \\ = 15- 6 \sqrt{2} + 5 \sqrt{2}-4 $$ Simplify denominator. $$ \color{blue}{ \left( 5 + \sqrt{8}\right) } \cdot \left( 5- \sqrt{8}\right) = \color{blue}{5} \cdot5+\color{blue}{5} \cdot- \sqrt{8}+\color{blue}{ \sqrt{8}} \cdot5+\color{blue}{ \sqrt{8}} \cdot- \sqrt{8} = \\ = 25- 10 \sqrt{2} + 10 \sqrt{2}-8 $$ |
| ③ | Simplify numerator and denominator |
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