$\frac{x^2+2}{x+3}>x$

$\left(1-\tan x\right)\left(\sin\left(2x\right)+1\right)=1+\tan\left(x\right)$

$\lim_{x\to0}\left(\frac{1}{\left(tan\left(x\right)^{-1}\right)^2}\right)$

$x^2\frac{dy}{dx}=\left(x+1\right)y$

$\frac{x^2+2x+1}{x^2+6x+7}$

$25+7\left(18-4^2\right)$

$6c=24$