$16x^2+40x-24$
$7\frac{3}{4}+3\frac{5}{6}-1\frac{3}{4}$
$\lim_{x\to0}\left(\frac{x^2-2x-3}{x-3}\right)$
$\frac{3x^6+x^4-x^2-3}{x+1}$
$x^{2+40x+400}$
$3\sqrt{2}-3\:.\sqrt{2}-1$
$6a^2\:-\:17a\:+\:12$
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