$6\:+\:12\:\cdot\:2\:-\:8$

$\lim_{x\to\infty}\left(\frac{x^3-27}{x^3+x}\right)^{x^2+1}$

$3x-8\ge11x+16$

$\frac{d}{dx}\left(\ln\left(\frac{\left(7x+2\right)^7}{\sqrt[5]{2x^2-10x+49}}\right)\right)$

$\sqrt[2]{\frac{s^2+1}{s^2+4}}$

$\frac{40}{\frac{\:2}{5}}$

$\frac{12\left(x-2\right)^2}{6\left(x-2\right)\left(x+5\right)}$