$\frac{x^3-3x^2-x+3}{x-5}$
$x^2+x-10$
$x^2-17x+10=0$
$\lim_{x\to3}\left(\frac{x^3-6x^2+9x}{\left(x-3\right)^2}\right)$
$-19\cdot37$
$\left(-4\right)\left(-18\right)\left(12\right)\left(15\right)\left(-19\right)$
$\int-12r^3\left(5-3r^4\right)^4dr$
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