Question
Answer
$$\dfrac{10+\sqrt{2}}{3-\sqrt{2}\cdot3}=-\dfrac{12+11\sqrt{2}}{3}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{10+\sqrt{2}}{3-\sqrt{2}\cdot3}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{10+\sqrt{2}}{3-\sqrt{2}\cdot3}\frac{3+3\sqrt{2}}{3+3\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{30+30\sqrt{2}+3\sqrt{2}+6}{9+9\sqrt{2}-9\sqrt{2}-18} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{36+33\sqrt{2}}{-9} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{12+11\sqrt{2}}{-3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}-\frac{12+11\sqrt{2}}{3}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 3 + 3 \sqrt{2}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 10 + \sqrt{2}\right) } \cdot \left( 3 + 3 \sqrt{2}\right) = \color{blue}{10} \cdot3+\color{blue}{10} \cdot 3 \sqrt{2}+\color{blue}{ \sqrt{2}} \cdot3+\color{blue}{ \sqrt{2}} \cdot 3 \sqrt{2} = \\ = 30 + 30 \sqrt{2} + 3 \sqrt{2} + 6 $$ Simplify denominator. $$ \color{blue}{ \left( 3- 3 \sqrt{2}\right) } \cdot \left( 3 + 3 \sqrt{2}\right) = \color{blue}{3} \cdot3+\color{blue}{3} \cdot 3 \sqrt{2}\color{blue}{- 3 \sqrt{2}} \cdot3\color{blue}{- 3 \sqrt{2}} \cdot 3 \sqrt{2} = \\ = 9 + 9 \sqrt{2}- 9 \sqrt{2}-18 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 3. |
| ⑤ | Place a negative sign in front of a fraction. |
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