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