$\frac{2yx^2}{4}$
$\frac{2z^2-8}{x^2+4x+4}$
$\left(-2+x\right)\left(a^2+b\right)$
$4<x\:+5$
$-2x-5=6-x-2x$
$4x-3y+10-2x-3y-6$
$3x\left(-14\right)$
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