$\frac{125m^6+0,729}{5m^2+0,9}$
$x^3-7x^2+25x-175\:$
$\left(\:x\:-\:15\:\right)\left(\:x\:-\:7\:\right)\:$
$-7\left(5+3\right)-\left|-6\right|$
$x^2+11x<18$
$\frac{6a^3b^2}{3a^2b-6ab^2}$
$\lim_{x\to-\infty}\left(1+\frac{2}{x^3}\right)x$
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