$-2\:\left|1-2\right|-\left|-3\right|$
$3x-2<8$
$4\left(2x-5\right)^3$
$3\left(5-3w\right)$
$-\frac{405}{15}$
$\frac{\left(2x^3-8x^2-1\right)}{x-4}$
$\tan^2\left(-x\right)+1=\sec^2\left(x\right)$
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