$x^5+4x^4+4x^3+2x^2+8x+8$
$3x\left(y^2+\:1\right)dx\:+\:y\left(x^2\:+\:2\right)dy\:=\:0$
$4a^3-2a^2$
$-3a^6b^2\left(-ab^3+ab+a^4b^6\right)$
$\sqrt[3]{\sqrt[4]{3x}}$
$\frac{1}{5}<\frac{x-3}{x+1}$
$\frac{dy}{dx}=\frac{\left(2y+x\right)}{x}$
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