$\left(4\right)+\left(-6\right)+\left(-8\right)+\left(10\right)+\left(2\right)$
$\left(\frac{1}{6}x^3-\frac{3}{2}x^2y^3\right)^3$
$\frac{1}{1-x}=\frac{2x}{3+x}$
$\frac{-5x^3-10x^2-5x-10}{-2x-1}$
$\frac{x-5}{3}+\frac{x+4}{2}>\frac{x+3}{6}$
$12x+32=12x-7$
$-2\cdot\left(-15\right)$
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