Question

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Explanation:

First Problem: $\boldsymbol{\frac{x}{x + 2} \div \frac{2}{x + 1}}$

Step1: Recall division of fractions rule

To divide two fractions, we multiply the first fraction by the reciprocal of the second fraction. So, $\frac{x}{x + 2} \div \frac{2}{x + 1}=\frac{x}{x + 2}\times\frac{x + 1}{2}$

Step2: Multiply numerators and denominators

Multiply the numerators $x$ and $x + 1$ and the denominators $x + 2$ and $2$. We get $\frac{x(x + 1)}{2(x + 2)}$

Step3: Expand the numerator (optional, but to simplify)

Expand $x(x + 1)$ to get $x^{2}+x$. So the simplified form is $\frac{x^{2}+x}{2x + 4}$

Step1: Factor all expressions

• Factor $5x + 5$: $5x+5 = 5(x + 1)$
• Factor $x^{2}-4x$: $x^{2}-4x=x(x - 4)$
• Factor $x^{2}-3x - 4$: We need two numbers that multiply to $- 4$ and add to $-3$. The numbers are $-4$ and $1$. So $x^{2}-3x - 4=(x - 4)(x+1)$

Step2: Apply division of fractions rule

$\frac{5(x + 1)}{2x}\div\frac{x(x - 4)}{(x - 4)(x + 1)}=\frac{5(x + 1)}{2x}\times\frac{(x - 4)(x + 1)}{x(x - 4)}$

Step3: Cancel out common factors

Cancel out $(x - 4)$ from numerator and denominator. We have $\frac{5(x + 1)}{2x}\times\frac{(x + 1)}{x}$

Step4: Multiply numerators and denominators

Multiply the numerators $5(x + 1)$ and $(x + 1)$ and the denominators $2x$ and $x$. We get $\frac{5(x + 1)^{2}}{2x^{2}}$

Step5: Expand the numerator (optional)

Expand $(x + 1)^{2}=x^{2}+2x + 1$, so $\frac{5(x^{2}+2x + 1)}{2x^{2}}=\frac{5x^{2}+10x + 5}{2x^{2}}$

Answer:

$\frac{x(x + 1)}{2(x + 2)}$ (or $\frac{x^{2}+x}{2x + 4}$)

Second Problem: $\boldsymbol{\frac{5x + 5}{2x} \div \frac{x^{2}-4x}{x^{2}-3x - 4}}$