$\int\:\frac{-3x}{\left(-12-6x\right)\left(3x-6\right)}dx$

$24>12\left(3-2x\right)$

$\left(4a^3-10\right)\left(4a^3-6\right)$

$-13am^3+11a^2m^2+10am^5$

$\lim\:_{x\to\infty\:}\left(\frac{0}{x}\right)$

$\int\frac{\sin\left(x\right)}{3}dx$

$0,25\:m+\frac{1}{\text{4}}m-2m$