Question
Answer
$$\dfrac{\sqrt{2}}{\sqrt{2}\cdot2-7\sqrt{3}}=-\dfrac{4+7\sqrt{6}}{139}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{2}}{\sqrt{2}\cdot2-7\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{2}}{\sqrt{2}\cdot2-7\sqrt{3}}\frac{2\sqrt{2}+7\sqrt{3}}{2\sqrt{2}+7\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{4+7\sqrt{6}}{8+14\sqrt{6}-14\sqrt{6}-147} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{4+7\sqrt{6}}{-139} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}-\frac{4+7\sqrt{6}}{139}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 2 \sqrt{2} + 7 \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \sqrt{2} } \cdot \left( 2 \sqrt{2} + 7 \sqrt{3}\right) = \color{blue}{ \sqrt{2}} \cdot 2 \sqrt{2}+\color{blue}{ \sqrt{2}} \cdot 7 \sqrt{3} = \\ = 4 + 7 \sqrt{6} $$ Simplify denominator. $$ \color{blue}{ \left( 2 \sqrt{2}- 7 \sqrt{3}\right) } \cdot \left( 2 \sqrt{2} + 7 \sqrt{3}\right) = \color{blue}{ 2 \sqrt{2}} \cdot 2 \sqrt{2}+\color{blue}{ 2 \sqrt{2}} \cdot 7 \sqrt{3}\color{blue}{- 7 \sqrt{3}} \cdot 2 \sqrt{2}\color{blue}{- 7 \sqrt{3}} \cdot 7 \sqrt{3} = \\ = 8 + 14 \sqrt{6}- 14 \sqrt{6}-147 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Place a negative sign in front of a fraction. |
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