$x^2+12x+40$
$\left(1\right)^{24}$
$\frac{3x}{28}=\frac{21}{4x}$
$2+5-1+5-2+2+3$
$\left(2a^5\:-\:162a^3\right)$
$2e\:\frac{3}{5\:}\:-\frac{7}{8}+1e\:\frac{1}{2}$
$\cos\left(x\right)=-0,4$
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