$\lim_{x\to\infty}\left(2xe^x\right)$
$\frac{5+3x}{23}<1$
$\lim_{x\to0}\left(\frac{x-\cos2x}{x-\sin4x}\right)$
$\frac{x^5-3x^4+9x^2+7x-4}{x^2-3x+2}$
$-4x^2-6x+4=0$
$\left(\frac{1}{2}m-\frac{2}{5}n\right).\left(\frac{1}{4}-\frac{2}{5}n\right)$
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