$\lim\:_{x\to\:0}\left(\frac{2^x-3^x}{3x}\right)$
$x^2-7x=12$
$-7x<42$
$3x=3y'-2$
$\frac{dy}{dx}=-y^2-y$
$y\:=\:2^{\left(2sen\:2\pi x\right)}3$
$\frac{x^2-2x-3}{\left(x+2\right)}+\frac{\left(3x-5\right)}{x+2}+\frac{6}{x+2}$
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