$\left(\frac{1}{2}x^3\right)\left(\frac{2}{3}a^2x\right)\left(-\frac{3}{5}a^4m\right)$
$\left(-35\right)\cdot\left(+10\right):\left(-7\right)\cdot4$
$m-48m+64$
$\left(tu\right)^{\frac{3}{4}}$
$\frac{0}{-15}$
$13a=13b$
$\frac{dy}{dx}=y^2+\frac{y}{x}-9\cdot x^2$
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