$-18\cdot8$
$h\left(t\right)=\frac{1}{1+1}$
$\lim_{x\to1}\left(\frac{x\:}{\ln\left(x\right)}\right)$
$3a^2x-9a^2$
$16m^2np+4mn^2+8mnp^2$
$\lim_{x\to2}\:\frac{2}{x-2}$
$20+9\left(-65\right)$
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