$x^2-4x+6<0$
$24y^2-48xy-18x^2$
$\left(x+5\right)^2+8\le0$
$y=x^2+4x;x\frac{dy}{dx}=\frac{2x}{3y^2}$
$\int\left(sin3x\right)\left(cos4x\right)dx$
$\sqrt[3]{u}=-5$
$\int x\cdot y^2dx$
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