$\lim_{x\to0}\left(x^2-5x\right)^{\frac{1}{x}}$
$\left(2x+5\right)\left(x^2-4x-1\right)$
$y^{'\:}+5=2x^2$
$2x^2+26x+72$
$\frac{2x+2h-1}{2x+2h+1}+\frac{2x+1}{2x-1}$
$y^4+2y^2+1=\left(y^2+1\right)^2$
$\left|-15\right|\left|-15\right|-\left|14\right|$
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