$\frac{4x^2+9x-4}{x+3}$
$\left(\frac{\left(x^2-9\right)}{x^2+5x-6}\right)\left(\frac{\left(x^2+x-2\right)}{x^2+4x-21}\right)$
$x^6-4x$
$5 x + 4 > 79$
$\infty^2-4\infty$
$-42+-4^2+2$
$m^5-10m;\:m=3$
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