Question
Answer
$$\dfrac{i}{(i-1)(i+2)(i+i)}=\dfrac{-3-i}{20}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{i}{(i-1)(i+2)(i+i)}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{i}{(1i^2+2i-i-2)\cdot2i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{i}{(1i^2+i-2)\cdot2i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{i}{(-1+i-2)\cdot2i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{i}{(1i-3)\cdot2i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}\frac{i}{2i^2-6i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle7}{\textcircled {7}} } }}}\frac{i}{-2-6i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle8}{\textcircled {8}} } }}}\frac{-3-i}{20}\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{i-1}\right) $ by each term in $ \left( i+2\right) $. $$ \left( \color{blue}{i-1}\right) \cdot \left( i+2\right) = i^2+2i-i-2 $$ |
| ② | Combine like terms: $$ \color{blue}{i} + \color{blue}{i} = \color{blue}{2i} $$ |
| ③ | Combine like terms: $$ i^2+ \color{blue}{2i} \color{blue}{-i} -2 = i^2+ \color{blue}{i} -2 $$ |
| ④ | $$ i^2 = -1 $$ |
| ⑤ | Combine like terms: $$ \color{blue}{-1} +i \color{blue}{-2} = i \color{blue}{-3} $$ |
| ⑥ | $$ \left( \color{blue}{i-3}\right) \cdot 2i = 2i^2-6i $$ |
| ⑦ | $$ 2i^2 = 2 \cdot (-1) = -2 $$ |
| ⑧ | Divide $ \, i \, $ by $ \, -2-6i \, $ to get $\,\, \dfrac{-3-i}{20} $. ( view steps ) |
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