$\frac{x^2+15}{\left(x+3\right)^2}$

$\frac{x}{6}-3=\frac{x}{6}$

$\lim_{x\to1}\left(\frac{x^2-1}{\left(x^2+1\right)\left(x-1\right)}\right)$

$3\cdot x^3\cdot y^{-5}$

$\frac{dr}{dt}=\frac{a}{s\left(t\right)}$

$\left(\frac{x^2}{3}+x-3\right)$

$-\left(5x-\frac{5}{7}\right)$