Question

⑦ $\frac{25 - x^{2}}{14x^{3}y^{8}}cdot\frac{7x^{2}y}{8x + 40}$

Explanation:

Step1: Factor the expressions

First, factor $25 - x^{2}$ using the difference - of - squares formula $a^{2}-b^{2}=(a + b)(a - b)$, so $25 - x^{2}=(5 + x)(5 - x)$. Factor out the greatest common factor from $8x + 40$, we get $8x+40 = 8(x + 5)$.

Step2: Rewrite the product of fractions

The original expression $\frac{25 - x^{2}}{14x^{3}y^{8}}\cdot\frac{7x^{2}y}{8x + 40}$ becomes $\frac{(5 + x)(5 - x)}{14x^{3}y^{8}}\cdot\frac{7x^{2}y}{8(x + 5)}$.

Step3: Simplify the coefficients and variables

Simplify the coefficients: $\frac{7}{14}=\frac{1}{2}$. For the $x$ - terms, use the rule $\frac{x^{2}}{x^{3}}=\frac{1}{x}$ (since $x^{m}\div x^{n}=x^{m - n}$). For the $y$ - terms, $\frac{y}{y^{8}}=\frac{1}{y^{7}}$ (using $y^{m}\div y^{n}=y^{m - n}$). Also, cancel out the common factor $(x + 5)$ in the numerator and denominator.
The simplified expression is $\frac{(5 - x)}{16xy^{7}}$.

Answer:

$\frac{5 - x}{16xy^{7}}$