$y'\:+\:xe^y\:=\:0$
$\frac{x^2+x-3}{1+x}$
$\sin\left(\frac{x}{3}\right)=\frac{1}{2}$
$\left(4a+3\right)\left(4a-8\right)$
$2x-3y+x-10y-b$
$y''+10y'+50y=0$
$\frac{6-3\left|2-\left(6-8\right)\right|^2}{-2\left|2-5\right|}$
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