$9p\:-\:5q\:+\:2p$
$\left(x^5+y^3\right)^2$
$-8j^4-3j^4$
$\left(h+7\right)\left(h-7\right)$
$\left(10-7-4\right)-\left(5-9+12\right)+\left(20-15\right)$
$\frac{x^2+4x+4}{x^2-1}$
$\lim_{x\to0}\left(\frac{12}{x}\right)$
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