Question
Answer
$$5\cdot\dfrac{i}{(1+i)\cdot(2+i)\cdot(3+i)}=\dfrac{1}{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}5 \cdot \frac{i}{(1+i)\cdot(2+i)\cdot(3+i)}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}5 \cdot \frac{i}{(2+i+2i+i^2)\cdot(3+i)} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}5 \cdot \frac{i}{(1i^2+3i+2)\cdot(3+i)} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}5 \cdot \frac{i}{(-1+3i+2)\cdot(3+i)} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}5 \cdot \frac{i}{(3i+1)\cdot(3+i)} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}5 \cdot \frac{i}{9i+3i^2+3+i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle6}{\textcircled {6}} } }}}5 \cdot \frac{i}{3i^2+10i+3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle7}{\textcircled {7}} } }}}5 \cdot \frac{i}{-3+10i+3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle8}{\textcircled {8}} } }}}5 \cdot \frac{i}{10i} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle9}{\textcircled {9}} } }}}5\cdot\frac{1}{10} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle10}{\textcircled {10}} } }}}\frac{1}{2}\end{aligned} $$ | |
| ① | Multiply each term of $ \left( \color{blue}{1+i}\right) $ by each term in $ \left( 2+i\right) $. $$ \left( \color{blue}{1+i}\right) \cdot \left( 2+i\right) = 2+i+2i+i^2 $$ |
| ② | Combine like terms: $$ 2+ \color{blue}{i} + \color{blue}{2i} +i^2 = i^2+ \color{blue}{3i} +2 $$ |
| ③ | $$ i^2 = -1 $$ |
| ④ | Combine like terms: $$ \color{blue}{-1} +3i+ \color{blue}{2} = 3i+ \color{blue}{1} $$ |
| ⑤ | Multiply each term of $ \left( \color{blue}{3i+1}\right) $ by each term in $ \left( 3+i\right) $. $$ \left( \color{blue}{3i+1}\right) \cdot \left( 3+i\right) = 9i+3i^2+3+i $$ |
| ⑥ | Combine like terms: $$ \color{blue}{9i} +3i^2+3+ \color{blue}{i} = 3i^2+ \color{blue}{10i} +3 $$ |
| ⑦ | $$ 3i^2 = 3 \cdot (-1) = -3 $$ |
| ⑧ | Combine like terms: $$ \, \color{blue}{ -\cancel{3}} \,+10i+ \, \color{blue}{ \cancel{3}} \, = 10i $$ |
| ⑨ | Divide $ \, i \, $ by $ \, 10i \, $ to get $\,\, \dfrac{1}{10} $. ( view steps ) |
| ⑩ | Multiply $5$ by $ \dfrac{1}{10} $ to get $ \dfrac{1}{2} $. Write $ 5 $ as a fraction by putting $ \color{red}{1} $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Cancel down by $ \color{blue}{5} $ $$ \begin{aligned} 5 \cdot \frac{1}{10} & = \frac{5}{\color{red}{1}} \cdot \frac{1}{10} = \frac{5 : \color{blue}{5}}{10 : \color{blue}{5}} = \\[1ex] &= \frac{1}{2} \end{aligned} $$ |
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