$\left(2x^2\right)\left(3x^3\right)\left(4x\right)^2$
$\frac{4a^2b^5}{64^3b^3}$
$\left(x-11\right)\left(x-4\right)$
$\int\:\sqrt{x}\left(x^2+1\right)dx$
$\frac{dy}{dx}\left(y^2+2y+2\right)$
$1,5\cdot23,99$
$-11-\left\{\left(+6\right)-\left(+4\right)\right\}$
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