$\frac{\left(x^3+3x^2+2\right)}{x+4}$
$2\left(3x-1\right)=4\left(x-3\right)$
$y'=\frac{5x}{y}$
$\int\frac{7x^4-3x^2+6}{x-7}dx$
$\frac{1}{\cos\left(x\right)\cdot\csc\left(x\right)}=\tan\left(x\right)$
$x^2y^6-36z^4$
$13\:-\:\left(9\:-\:16\right)$
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