$10x^2-40x=-40$
$\left(\frac{4x-5}{3}\right)=\left(\frac{3x-7}{2}\right)$
$\left(\left(4\cdot12\right)\left(4\cdot8\right)\right)$
$y'=2^{x-y}$
$\left(\frac{3}{2}x^2-\frac{1}{4}y^3\right)^2$
$\frac{dy}{dx}=\frac{3y^2}{x}$
$\int_0^412xdx$
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