Question
Answer
$$\dfrac{(5\sqrt{2}+2\sqrt{3})\cdot2}{3+2\sqrt{2}}=30\sqrt{2}-40+12\sqrt{3}-8\sqrt{6}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{(5\sqrt{2}+2\sqrt{3})\cdot2}{3+2\sqrt{2}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{10\sqrt{2}+4\sqrt{3}}{3+2\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{10\sqrt{2}+4\sqrt{3}}{3+2\sqrt{2}}\frac{3-2\sqrt{2}}{3-2\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{30\sqrt{2}-40+12\sqrt{3}-8\sqrt{6}}{9-6\sqrt{2}+6\sqrt{2}-8} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{30\sqrt{2}-40+12\sqrt{3}-8\sqrt{6}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}30\sqrt{2}-40+12\sqrt{3}-8\sqrt{6}\end{aligned} $$ | |
| ① | $$ \color{blue}{ \left( 5 \sqrt{2} + 2 \sqrt{3}\right) } \cdot 2 = \color{blue}{ 5 \sqrt{2}} \cdot2+\color{blue}{ 2 \sqrt{3}} \cdot2 = \\ = 10 \sqrt{2} + 4 \sqrt{3} $$ |
| ② | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 3- 2 \sqrt{2}} $$. |
| ③ | Multiply in a numerator. $$ \color{blue}{ \left( 10 \sqrt{2} + 4 \sqrt{3}\right) } \cdot \left( 3- 2 \sqrt{2}\right) = \color{blue}{ 10 \sqrt{2}} \cdot3+\color{blue}{ 10 \sqrt{2}} \cdot- 2 \sqrt{2}+\color{blue}{ 4 \sqrt{3}} \cdot3+\color{blue}{ 4 \sqrt{3}} \cdot- 2 \sqrt{2} = \\ = 30 \sqrt{2}-40 + 12 \sqrt{3}- 8 \sqrt{6} $$ Simplify denominator. $$ \color{blue}{ \left( 3 + 2 \sqrt{2}\right) } \cdot \left( 3- 2 \sqrt{2}\right) = \color{blue}{3} \cdot3+\color{blue}{3} \cdot- 2 \sqrt{2}+\color{blue}{ 2 \sqrt{2}} \cdot3+\color{blue}{ 2 \sqrt{2}} \cdot- 2 \sqrt{2} = \\ = 9- 6 \sqrt{2} + 6 \sqrt{2}-8 $$ |
| ④ | Simplify numerator and denominator |
| ⑤ | Remove 1 from denominator. |
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