$2x=x+12$
$x+\frac{1}{x^3}$
$\frac{2}{3}x^{\frac{3}{2}}-\frac{1}{2}x^{\frac{1}{2}}$
$x^2-10>0$
$x^{n-1}.\:1$
$2\left(x-3\right)\ge4\left(x+4\right)-10$
$\lim_{x\to\infty}\left(\frac{\sin\left(2x\right)}{x}\right)$
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