Answer
$$\frac{\sqrt{3}-2}{\sqrt{31}+\sqrt{6}\cdot\sqrt{4}\cdot15+\sqrt{-4}\cdot\sqrt{4}\cdot3}=\frac{\sqrt{3}-2}{\sqrt{31}+30\sqrt{6}+12i}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{3}-2}{\sqrt{31}+\sqrt{6}\cdot\sqrt{4}\cdot15+\sqrt{-4}\cdot\sqrt{4}\cdot3}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{3}-2}{\sqrt{31}+30\sqrt{6}+\sqrt{4}\cdot\sqrt{4}i\cdot3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{\sqrt{3}-2}{\sqrt{31}+30\sqrt{6}+2\cdot2i\cdot3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{\sqrt{3}-2}{\sqrt{31}+30\sqrt{6}+4i\cdot3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{\sqrt{3}-2}{\sqrt{31}+30\sqrt{6}+12i}\end{aligned} $$ | |
| ① | $$ 15 \sqrt{24} =
15 \sqrt{ 2 ^2 \cdot 6 } =
15 \sqrt{ 2 ^2 } \, \sqrt{ 6 } =
15 \cdot 2 \sqrt{ 6 } =
30 \sqrt{ 6 } $$ |
| ② | $$ 15 \sqrt{24} =
15 \sqrt{ 2 ^2 \cdot 6 } =
15 \sqrt{ 2 ^2 } \, \sqrt{ 6 } =
15 \cdot 2 \sqrt{ 6 } =
30 \sqrt{ 6 } $$ |
| ③ | $$ 15 \sqrt{24} =
15 \sqrt{ 2 ^2 \cdot 6 } =
15 \sqrt{ 2 ^2 } \, \sqrt{ 6 } =
15 \cdot 2 \sqrt{ 6 } =
30 \sqrt{ 6 } $$ |
| ④ | $$ 4 i \cdot 3 = 12 i $$ |
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