$3 x + 1 \leq 12$
$y\res\tan-\frac{1}{12}x^{2}y$
$4\left(a+b\right)$
$\frac{x^3+4x^2-3x-18}{x-4}$
$12+x-x^2\ge0$
$\left(a^2b-6b\right)\left(3a^2b+3b\right)$
$4w^2+72w+324$
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