$3\cdot2-5+4\cdot3-8+5\cdot2$
$\lim_{x\to\infty}\left(\frac{x^3-4x^2+8}{2x-6x^2}\right)$
$3xy^{2\:}+4x^3y^2+6xy$
$y=3\left(2x-1\right)\left(x+4\right)$
$16+625x^2+300x+x+180$
$\sqrt[60]{\frac{5474.94}{4500}}-1$
$3\left(2x-5\right)+8x-6\ge\frac{x}{2}-\left(5x+3\right)$
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