Question
Answer
$$\dfrac{1+\sqrt{3}}{2+\sqrt{2}}=\dfrac{2-\sqrt{2}+2\sqrt{3}-\sqrt{6}}{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{1+\sqrt{3}}{2+\sqrt{2}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{1+\sqrt{3}}{2+\sqrt{2}}\frac{2-\sqrt{2}}{2-\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{2-\sqrt{2}+2\sqrt{3}-\sqrt{6}}{4-2\sqrt{2}+2\sqrt{2}-2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{2-\sqrt{2}+2\sqrt{3}-\sqrt{6}}{2}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 2- \sqrt{2}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 1 + \sqrt{3}\right) } \cdot \left( 2- \sqrt{2}\right) = \color{blue}{1} \cdot2+\color{blue}{1} \cdot- \sqrt{2}+\color{blue}{ \sqrt{3}} \cdot2+\color{blue}{ \sqrt{3}} \cdot- \sqrt{2} = \\ = 2- \sqrt{2} + 2 \sqrt{3}- \sqrt{6} $$ Simplify denominator. $$ \color{blue}{ \left( 2 + \sqrt{2}\right) } \cdot \left( 2- \sqrt{2}\right) = \color{blue}{2} \cdot2+\color{blue}{2} \cdot- \sqrt{2}+\color{blue}{ \sqrt{2}} \cdot2+\color{blue}{ \sqrt{2}} \cdot- \sqrt{2} = \\ = 4- 2 \sqrt{2} + 2 \sqrt{2}-2 $$ |
| ③ | Simplify numerator and denominator |
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