Question
Answer
$$\dfrac{2\sqrt{2}-2}{3\sqrt{2}+1}=\dfrac{14-8\sqrt{2}}{17}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2\sqrt{2}-2}{3\sqrt{2}+1}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2\sqrt{2}-2}{3\sqrt{2}+1}\frac{3\sqrt{2}-1}{3\sqrt{2}-1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{12-2\sqrt{2}-6\sqrt{2}+2}{18-3\sqrt{2}+3\sqrt{2}-1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{14-8\sqrt{2}}{17}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 3 \sqrt{2}-1} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 2 \sqrt{2}-2\right) } \cdot \left( 3 \sqrt{2}-1\right) = \color{blue}{ 2 \sqrt{2}} \cdot 3 \sqrt{2}+\color{blue}{ 2 \sqrt{2}} \cdot-1\color{blue}{-2} \cdot 3 \sqrt{2}\color{blue}{-2} \cdot-1 = \\ = 12- 2 \sqrt{2}- 6 \sqrt{2} + 2 $$ Simplify denominator. $$ \color{blue}{ \left( 3 \sqrt{2} + 1\right) } \cdot \left( 3 \sqrt{2}-1\right) = \color{blue}{ 3 \sqrt{2}} \cdot 3 \sqrt{2}+\color{blue}{ 3 \sqrt{2}} \cdot-1+\color{blue}{1} \cdot 3 \sqrt{2}+\color{blue}{1} \cdot-1 = \\ = 18- 3 \sqrt{2} + 3 \sqrt{2}-1 $$ |
| ③ | Simplify numerator and denominator |
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