$2x^2-3x-18=0$
$\lim_{x\to\infty}\left(\frac{1-3x^2}{x^2+1}\right)$
$\frac{2x+1}{5}<\frac{3x+4}{10}$
$2x^2-6x+3$
$25x\cdot5x$
$n+5+3-2n$
$\frac{\sin\left(a\right)}{\cos\left(a\right)-1}=-\cot\left(\frac{a}{2}\right)$
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