Question
Answer
$$-x\cdot\dfrac{i}{1-xi}=-\dfrac{ix}{-ix+1}$$
Explanation
| $$ \begin{aligned}-x \cdot \frac{i}{1-xi}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}-\frac{ix}{-ix+1}\end{aligned} $$ | |
| ① | Multiply $x$ by $ \dfrac{i}{1-ix} $ to get $ \dfrac{ix}{-ix+1} $. Step 1: Write $ x $ as a fraction by putting $ \color{red}{1} $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} x \cdot \frac{i}{1-ix} & \xlongequal{\text{Step 1}} \frac{x}{\color{red}{1}} \cdot \frac{i}{1-ix} \xlongequal{\text{Step 2}} \frac{ x \cdot i }{ 1 \cdot \left( 1-ix \right) } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ ix }{ 1-ix } = \frac{ix}{-ix+1} \end{aligned} $$ |
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