$6\left(1\right)\frac{3}{2}$
$subtract\:65\:from\:\left(-80\right)$
$\int\left(4x^3-2\right)^3dx$
$0.2\left(-3\right)^2-3.5\left(-2\right)$
$8-2-4+6-3+5$
$\frac{d}{dx}x^2+4y^2-2x+y=0$
$\frac{4x^{4}+9x^{3}+6x^{5}-1}{x+2x^{3}-1}$
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