$6\:4\:x\:4\:4\:-2\:5\:y\:2\:$
$\int\frac{\left(x^2-5x+6\right)}{x-3}dx$
$\frac{dy}{dx}=-sin\:2x$
$x^2-x+0.25$
$\frac{x+4}{9}=\frac{x-2}{7}$
$\left(3x^2+4z^3\right)^2$
$\frac{x^2-16}{x^2+4}$
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