$4-3^2+4$
$2+3\cdot\left(-4+2\cdot7\right)-\left(-9+8-6\right)$
$12\cdot1$
$81z^2+36z+4$
$\frac{3-x}{x^2+2}-\frac{2x-3}{2x+4}$
$\left(4x^3-12x\right)$
$\lim_{x\to4}\left(\frac{\left|x^2-16\right|+\left|x-4\right|}{x-4}\right)$
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