$x^2+6y-1;\:x=-3$
$y'=x\left(-8+y\right)$
$\frac{7v^2}{4v^3}$
$\left(x-1\right)\left(x^2+x+1.\right)$
$\left(3\right)^3-\left(1\right)^3$
$4x>\frac{1}{6}\left(x-2\right)+3$
$\frac{\left(5a^2x^4\right)}{-3x^2}$
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