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

$\lim_{x\to0}\left(\frac{6-6\cdot cos\left(x\right)}{sin\left(7x\right)}\right)$

$\lim_{x\to2}\left(\frac{x^3-8}{x^2-2x}\right)$

$f\left(x\right)=\left(x^3-2x+1\right)\left(2x^2+3x\right)$

$\frac{2}{3}m^6n^3\cdot\frac{5}{8}m^5n^2$

$x^2-4>4$

$6r^2\:-\:5r^5\:-\:5r^4\:+\:12r$