$-3\cos\left(x\right)+3=\cos\left(x\right)+5$

$-10.9p\:+\:3.9\:=\:-9.18$

$\lim_{k\to3}\frac{k^2-7k+12}{k-3}$

$2cosx-3tanx=0$

$\lim_{x\to0}\left(3x^2-2x+5\right)$

$\frac{\left(2\cdot3^{-2}\right)^{3}\left(5\cdot3^{2}\right)^{2}}{\left(3^{-2}\right)\left(5\cdot2\right)^{2}}$

$2r^2-12r+18$