Question
Answer
$$\dfrac{(5+\sqrt{2})\cdot(5-\sqrt{2})}{\sqrt{2}\cdot3}=\dfrac{23\sqrt{2}}{6}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{(5+\sqrt{2})\cdot(5-\sqrt{2})}{\sqrt{2}\cdot3}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{25-5\sqrt{2}+5\sqrt{2}-2}{\sqrt{2}\cdot3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{23}{3\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{23}{3\sqrt{2}}\frac{\sqrt{2}}{\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{23\sqrt{2}}{6}\end{aligned} $$ | |
| ① | $$ \color{blue}{ \left( 5 + \sqrt{2}\right) } \cdot \left( 5- \sqrt{2}\right) = \color{blue}{5} \cdot5+\color{blue}{5} \cdot- \sqrt{2}+\color{blue}{ \sqrt{2}} \cdot5+\color{blue}{ \sqrt{2}} \cdot- \sqrt{2} = \\ = 25- 5 \sqrt{2} + 5 \sqrt{2}-2 $$ |
| ② | Simplify numerator and denominator |
| ③ | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{2}} $$. |
| ④ | Multiply in a numerator. $$ \color{blue}{ 23 } \cdot \sqrt{2} = 23 \sqrt{2} $$ Simplify denominator. $$ \color{blue}{ 3 \sqrt{2} } \cdot \sqrt{2} = 6 $$ |
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