$x+8>17$
$\int\frac{5x^4-x^2}{3x^5-x^3}\:dx$
$\lim_{g\to0}\left(3x-2\right)$
$\frac{x\:^5+\:3x}{x}$
$\int\frac{\left(5x^2-3x+9\right)}{\left(x-3\right)\left(x^2+4\right)}dx$
$2x+6<-2$
$-4x^2y+8xy\cdot xy$
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