$\lim_{x\to-1}\left(\frac{x^2-4x}{x^2-3x-4}\:\:\:\right)$
$x^2-18x-91x$
$\left(-10x+4\right)^2$
$3x^2\left(1+lny\right)dx+\left(\frac{x^3}{y}-8y\right)dy=0$
$\frac{x^5}{1+x^3}$
$\frac{125x^9+x^{12}}{5x^3+x^4}$
$l^2mas\:l^2$
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