Answer
$$\frac{1+\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}}=-\sqrt{2}-1$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{1+\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\frac{2+\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}} \xlongequal{ } \\[1 em] & \xlongequal{ }\frac{2+\sqrt{2}}{2}\cdot\frac{2}{-\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{4+2\sqrt{2}}{-2\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{4+2\sqrt{2}}{-2\sqrt{2}}\frac{\sqrt{2}}{\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{4\sqrt{2}+4}{-4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{\sqrt{2}+1}{-1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}-\frac{\sqrt{2}+1}{1} \xlongequal{ } \\[1 em] & \xlongequal{ }-(\sqrt{2}+1) \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle7}{\textcircled {7}} } }}}-\sqrt{2}-1\end{aligned} $$ | |
| ① | $$ 1+\frac{\sqrt{2}}{2}
= 1 \cdot \color{blue}{\frac{ 2 }{ 2}} + \frac{\sqrt{2}}{2} \cdot \color{blue}{\frac{ 1 }{ 1}}
= \frac{2+\sqrt{2}}{2} $$ |
| ② | $$ \color{blue}{ \left( 2 + \sqrt{2}\right) } \cdot 2 = \color{blue}{2} \cdot2+\color{blue}{ \sqrt{2}} \cdot2 = \\ = 4 + 2 \sqrt{2} $$$$ \color{blue}{ 2 } \cdot - \sqrt{2} = - 2 \sqrt{2} $$ |
| ③ | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{2}} $$. |
| ④ | Multiply in a numerator. $$ \color{blue}{ \left( 4 + 2 \sqrt{2}\right) } \cdot \sqrt{2} = \color{blue}{4} \cdot \sqrt{2}+\color{blue}{ 2 \sqrt{2}} \cdot \sqrt{2} = \\ = 4 \sqrt{2} + 4 $$ Simplify denominator. $$ \color{blue}{ - 2 \sqrt{2} } \cdot \sqrt{2} = -4 $$ |
| ⑤ | Divide both numerator and denominator by 4. |
| ⑥ | Place a negative sign in front of a fraction. |
| ⑦ | Remove the parenthesis by changing the sign of each term within them. |
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