$5x\left(2x^2-3x+4\right)$
$y^{3}=s\left(3x-t\right)$
$\frac{d}{dx}xy=2x^2+3y^2$
$\frac{157}{25}\cdot\left(x+y\right)$
$\left(3x^{2}-5x-2\right)\left(2x^{2}-7x+4\right)$
$^{x^{2019}+6=\left(e\right)^x}$
$x^2+y^2+y''=36+x+y''$
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