Question
Answer
$$\dfrac{\sqrt{3}+2}{1+\sqrt{2}}=-\sqrt{3}+\sqrt{6}-2+2\sqrt{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{3}+2}{1+\sqrt{2}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{3}+2}{1+\sqrt{2}}\frac{1-\sqrt{2}}{1-\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{\sqrt{3}-\sqrt{6}+2-2\sqrt{2}}{1-\sqrt{2}+\sqrt{2}-2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{\sqrt{3}-\sqrt{6}+2-2\sqrt{2}}{-1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-\sqrt{3}+\sqrt{6}-2+2\sqrt{2}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}-\sqrt{3}+\sqrt{6}-2+2\sqrt{2}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 1- \sqrt{2}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{3} + 2\right) } \cdot \left( 1- \sqrt{2}\right) = \color{blue}{ \sqrt{3}} \cdot1+\color{blue}{ \sqrt{3}} \cdot- \sqrt{2}+\color{blue}{2} \cdot1+\color{blue}{2} \cdot- \sqrt{2} = \\ = \sqrt{3}- \sqrt{6} + 2- 2 \sqrt{2} $$ Simplify denominator. $$ \color{blue}{ \left( 1 + \sqrt{2}\right) } \cdot \left( 1- \sqrt{2}\right) = \color{blue}{1} \cdot1+\color{blue}{1} \cdot- \sqrt{2}+\color{blue}{ \sqrt{2}} \cdot1+\color{blue}{ \sqrt{2}} \cdot- \sqrt{2} = \\ = 1- \sqrt{2} + \sqrt{2}-2 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
| ⑤ | Remove 1 from denominator. |
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