$13x^2-12x+3=0$
$ycos\left(x\right)dx+dy=0$
$\left(4x^2+\:x\right)$
$\left|-16-\left(-5\right)\right|+\left(-2\right)\cdot6-8$
$\frac{x^4-2x^3+3\:x-6}{x-3}$
$\left(5x^2+3y^2\right)\left(4x^2-2y^3\right)$
$\frac{x}{x^2-2}=\frac{-1}{x}$
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