$-0,55-\:-1.2$
$\int\frac{1}{\left(4+x\right)^2\left(4+3x\right)}dx$
$\int\frac{2x^4-6x^3+5}{x-2}dx$
$\lim_{x\to2}\left(4x-3\right)$
$x^2-3x-1=0$
$\frac{3}{m^{2}}\cdot\frac{16m}{a}$
$8 x ^ { 3 } + 4 x ^ { 2 } + 2 x - 4$
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