$9m-m^2$
$\lim_{x\to0}\left(\frac{\sin3x}{x\cos2x}\right)$
$9\cdot\left(5\right)^2$
$2\cdot18$
$\frac{1}{6}tan^23y$
$\left(-32\right)-\left(-18\right)$
$\left(6c+5f\right)\left(6c-5f\right)$
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