$\frac{d^2}{dx^2}\left(\left(2x^2-4x+1\right)\left(6x-5\right)\right)$
$5+\sqrt{x+1}=x+2$
$\frac{\left(3-4\right)^2}{3-3}$
$8t^3-64$
$\int19\sin^2\left(x\right)\cos^5\left(x\right)dx$
$\sqrt[4]{0.4}$
$4\left(4-3\right)$
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