$3x-4=5$
$\frac{3x^3-2x^2+3x-1}{x+1}$
$\lim_{x\to0}\left(\frac{x^3+4x^2+4x}{x^2+4x}\right)$
$45\cdot101$
$\frac{dy}{dx}=\left(y-4x\right)^2$
$\lim_{x\to0}\left(\left(\frac{e^{2x}-1}{3x^2+2x}\right)\right)$
$-7x^2+12x+36$
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