$\:\left(x^2+4\right)\left(x+4\right)\left(x^4+2x^2-8\right)=0$
$\frac{d}{dx}3y$
$20x^4y^3-10x^3y^4-15x^2y^5$
$0,\:5y''+3y'+9y=0$
$\frac{3}{x^2-2x+1}+\frac{2}{x^2-1}$
$8x-2y-8$
$9x\:+\:5\:=\:-13$
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