Answer
a. $\frac{7}{x}$
b. $\frac{(x+3)^2}{7x}$
c. $-\frac{2}{(x+1)(x-4)}$
Work Step by Step
a.
$\frac{x+3}{x} \times \frac{7}{x+3}$
$\frac{(x+3)7}{x(x+3)}$
$=\frac{7}{x}$
b.
$\frac{x+3}{x} \div \frac{7}{x+3}$
$=\frac{x+3}{x} \times \frac{x+3}{7}$
$=\frac{(x+3)(x+3)}{x(7)}$
$=\frac{(x+3)^2}{7x}$
c.
$\frac{3-x}{x^2+6x+5} \times \frac{2x+10}{x^2-7x+12}$
$=\frac{3-x}{(x+1)(x+5)} \times \frac{2(x+5)}{(x-3)(x-4)}$
$=\frac{-(x-3)}{(x+1)(x+5)} \times \frac{2(x+5)}{(x-3)(x-4)}$
$=\frac{-1}{(x+1)} \times \frac{2}{(x-4)}$
$=-\frac{2}{(x+1)(x-4)}$
