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

$\frac{\left(x+2\right)}{\left(x+2\right)^2}\cdot\frac{\left(x^2-4\right)}{x}$

$1.4x-2.9y+4.8x-0.3y$

$\lim_{x\to3}\left(\frac{x^2-4x\:+3}{x^2+9}\right)$

$2cos3x=1$

$\tan u\:\csc^2u-\cot\:u\:\sec^2\:u=0$

$\left(+3\right)-\left(-4\right)+\left(-5\right)-\left(+8\right)$