Question
Answer
$$\dfrac{\dfrac{iwl}{irwc-w^2lc}}{iwc}=\dfrac{ilw}{c^2i^2rw^2-c^2ilw^3}$$
Explanation
| $$ \begin{aligned}\frac{\frac{iwl}{irwc-w^2lc}}{iwc}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{ilw}{c^2i^2rw^2-c^2ilw^3}\end{aligned} $$ | |
| ① | Divide $ \dfrac{ilw}{cirw-clw^2} $ by $ ciw $ to get $ \dfrac{ ilw }{ c^2i^2rw^2-c^2ilw^3 } $. Step 1: To divide rational expressions, multiply the first fraction by the reciprocal of the second fraction. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} \frac{ \frac{ilw}{cirw-clw^2} }{ciw} & \xlongequal{\text{Step 1}} \frac{ilw}{cirw-clw^2} \cdot \frac{\color{blue}{1}}{\color{blue}{ciw}} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ ilw \cdot 1 }{ \left( cirw-clw^2 \right) \cdot ciw } \xlongequal{\text{Step 3}} \frac{ ilw }{ c^2i^2rw^2-c^2ilw^3 } \end{aligned} $$ |
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