$\left(4^{-3}\right)^2$
$2x^5-4x^3+8x^2-6x$
$\lim_{x\to0}\left(\frac{sin\left(9x\right)}{tan\left(3x\right)}\right)$
$\frac{2x}{x+3}\ge1$
$6x^2-9x=0$
$-m-2m-8m-4m-m$
$5sin\left(2x+\frac{\pi\:}{3}\right)$
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