$\left(1-6\right)^2-8$
$\lim_{x\to\infty}\left(\frac{\left(-e\right)^x}{4}\right)$
$3x-1\le1-4\left(x-1\right)$
$m^3+15m^3+75m+125$
$\frac{d^2}{dx^2}\left(4x^2\:+\:6x\:-\:12\right)$
$5x-7\le-12$
$25x^2-6x+8$
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