$-1+-1+15$
$0.72\cdot1.27$
$3x\:-\:2\:>=\:4x\:+\:8$
$\lim_{x\to0}\left(2x^{3\:\:}\right)^{\frac{1}{x}}$
$-2>\frac{-7-1}{2}>-4$
$4\cdot\left(-8\right)-\left(+9\right)\cdot4$
$\frac{\left(3x+x^2\right)}{1}^2$
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