$\left(2y^3-3yx^2\right)dx+2x^3dy=0$
$\left(2x+1\right)\ge3$
$-0,3-0,7$
$\left(\frac{2}{x+3}\right)\left(\frac{x}{x+3}\right)$
$\frac{6-y}{2}=14-\frac{14-3y}{4}$
$40=10\log\left(\frac{5}{x^2}\right)$
$4\frac{-3n}{3}=-n$
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