$\frac{12x^8y^4}{x^3y^5}$
$\frac{\left(2\cdot3\cdot5\right)^{3}}{15^{2}\cdot60}$
$2x+5>4x-8$
$\left(-x^2-y^2\right)^3$
$12x+20y-15;\:x=4;\:y=5$
$8x+5y+6=0$
$x^3-y^3=2x$
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