$\lim_{x\to\infty}\left(\frac{x+1}{3}\right)$
$3x\:-9y\:-5y\:+8x$
$\int\:x^3\left(x^2+1\right)^8dx$
$13+2\left(x-6\right)=4x-7$
$\left(4^1\right)\left(4^5\right)$
$5y\:-\:12\:=73$
$\frac{d}{dx}-4y^2-9=0$
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