$\lim_{t\to-2}\left(\sqrt{8+t^3}\right)$
$\frac{dx}{y}+\left(x^2-4x\right)dy=0$
$\frac{dy}{dx}=\frac{6y^3+6x^3}{6xy^2}$
$6x^2+15x-36$
$x^4y^2-x^2y^4=0$
$-9x-6\ge4x+13$
$\left(2-x\right)\sqrt{x}$
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