$3-x<2+5x$
$v^2-9v-8$
$18-15-12-36+95-5+18$
$\frac{1}{x-1}=\frac{1}{x+1}$
$\frac{\left(-1\right)^nx^{n+1}}{5^n}$
$-6x^5\cdot3x^3$
$\frac{-1}{infinito}$
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