$\int_0^{2\sqrt{2}}\left(\frac{x^2}{\sqrt{16-x^2}}\right)dx$
$16x^6y^2+z^2-8x^3yz$
$\left(\frac{x}{2}-\frac{y}{3}\right)^4$
$\frac{dy}{dx}=\frac{\left(x-y\right)}{\left(x+y\right)}$
$\frac{4x^3-3x^2}{2x^2}$
$8\:x\:-\:288\:=\:0\:$
$x^2-6x+20$
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