$\frac{x}{3}+\frac{x}{2}5-\frac{x}{6}$

$\lim_{x\to6}\left(\frac{e^{-6x}}{4+e^{-3x}}\right)$

$x^4-6x^3-7x^2+60x=0$

$\int\frac{1}{\left(x^2-16\right)^2}dx$

$\left(-2x\right).\left(-x^2-1\right)$

$\frac{-\frac{1}{3}x-\frac{1}{4}}{\frac{1}{2}}\le\frac{3}{2}+1$

$\lim_{x\to1\:}2x^4$