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Answer

$$\frac{1-\frac{4}{x}}{1-\frac{16}{x^2}}=\frac{x^2}{x^2+4x}$$

Explanation

Tap the blue circles to see an explanation.

$$ \begin{aligned}\frac{1-\frac{4}{x}}{1-\frac{16}{x^2}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{\frac{x-4}{x}}{\frac{x^2-16}{x^2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{x^2}{x^2+4x}\end{aligned} $$
①

Subtract $ \dfrac{4}{x} $ from $ 1 $ to get $ \dfrac{ \color{purple}{ x-4 } }{ x }$.

Step 1: Write $ 1 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: To subtract raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $\color{blue}{ x }$.

$$ \begin{aligned} 1- \frac{4}{x} & \xlongequal{\text{Step 1}} \frac{1}{\color{red}{1}} - \frac{4}{x} = \frac{ 1 \cdot \color{blue}{ x }}{ 1 \cdot \color{blue}{ x }} - \frac{ 4 }{ x } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ x } }{ x } - \frac{ \color{purple}{ 4 } }{ x }=\frac{ \color{purple}{ x-4 } }{ x } \end{aligned} $$
②

Subtract $ \dfrac{16}{x^2} $ from $ 1 $ to get $ \dfrac{ \color{purple}{ x^2-16 } }{ x^2 }$.

Step 1: Write $ 1 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: To subtract raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $\color{blue}{ x^2 }$.

$$ \begin{aligned} 1- \frac{16}{x^2} & \xlongequal{\text{Step 1}} \frac{1}{\color{red}{1}} - \frac{16}{x^2} = \frac{ 1 \cdot \color{blue}{ x^2 }}{ 1 \cdot \color{blue}{ x^2 }} - \frac{ 16 }{ x^2 } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ x^2 } }{ x^2 } - \frac{ \color{purple}{ 16 } }{ x^2 }=\frac{ \color{purple}{ x^2-16 } }{ x^2 } \end{aligned} $$
③

Divide $ \dfrac{x-4}{x} $ by $ \dfrac{x^2-16}{x^2} $ to get $ \dfrac{ x^2 }{ x^2+4x } $.

Step 1: To divide rational expressions, multiply the first fraction by the reciprocal of the second fraction.

Step 2: Factor numerators and denominators.

Step 3: Cancel common factors.

Step 4: Multiply numerators and denominators.

Step 5: Simplify numerator and denominator.

$$ \begin{aligned} \frac{ \frac{x-4}{x} }{ \frac{\color{blue}{x^2-16}}{\color{blue}{x^2}} } & \xlongequal{\text{Step 1}} \frac{x-4}{x} \cdot \frac{\color{blue}{x^2}}{\color{blue}{x^2-16}} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ 1 \cdot \color{blue}{ \left( x-4 \right) } }{ x } \cdot \frac{ x^2 }{ \left( x+4 \right) \cdot \color{blue}{ \left( x-4 \right) } } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 1 }{ x } \cdot \frac{ x^2 }{ x+4 } \xlongequal{\text{Step 4}} \frac{ 1 \cdot x^2 }{ x \cdot \left( x+4 \right) } \xlongequal{\text{Step 5}} \frac{ x^2 }{ x^2+4x } \end{aligned} $$
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