Question
Answer
$$\dfrac{2\sqrt{2}-3\sqrt{2}}{2\sqrt{3}+2\sqrt{2}}=\dfrac{-\sqrt{6}+2}{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2\sqrt{2}-3\sqrt{2}}{2\sqrt{3}+2\sqrt{2}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{-\sqrt{2}}{2\sqrt{3}+2\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{-\sqrt{2}}{2\sqrt{3}+2\sqrt{2}}\frac{2\sqrt{3}-2\sqrt{2}}{2\sqrt{3}-2\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{-2\sqrt{6}+4}{12-4\sqrt{6}+4\sqrt{6}-8} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-2\sqrt{6}+4}{4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{-\sqrt{6}+2}{2}\end{aligned} $$ | |
| ① | Simplify numerator and denominator |
| ② | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 2 \sqrt{3}- 2 \sqrt{2}} $$. |
| ③ | Multiply in a numerator. $$ \color{blue}{ - \sqrt{2} } \cdot \left( 2 \sqrt{3}- 2 \sqrt{2}\right) = \color{blue}{- \sqrt{2}} \cdot 2 \sqrt{3}\color{blue}{- \sqrt{2}} \cdot- 2 \sqrt{2} = \\ = - 2 \sqrt{6} + 4 $$ Simplify denominator. $$ \color{blue}{ \left( 2 \sqrt{3} + 2 \sqrt{2}\right) } \cdot \left( 2 \sqrt{3}- 2 \sqrt{2}\right) = \color{blue}{ 2 \sqrt{3}} \cdot 2 \sqrt{3}+\color{blue}{ 2 \sqrt{3}} \cdot- 2 \sqrt{2}+\color{blue}{ 2 \sqrt{2}} \cdot 2 \sqrt{3}+\color{blue}{ 2 \sqrt{2}} \cdot- 2 \sqrt{2} = \\ = 12- 4 \sqrt{6} + 4 \sqrt{6}-8 $$ |
| ④ | Simplify numerator and denominator |
| ⑤ | Divide both numerator and denominator by 2. |
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