$4x\:-\:3\:+5x\:-12x\:+8$
$0p\left(-7\right)$
$2.2^3-2^2+6$
$4+x\le2+3x$
$\lim_{x\to0}\left(\frac{\sin\left(2x\right)}{x^2\sin\left(x\right)}-\frac{2}{x^2}\right)$
$-\left(-18\right):\:\left(-2\right)-\left(5-10\right)$
$\frac{2^8}{2^3}$
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