$\left(6\frac{1}{3}-2\frac{1}{4}\right)\left(3-2^2-\frac{1}{2}^3\right)$
$\frac{\left(3a^2b^4c^2\right)^3}{9a^5b^7c^6}$
$\cot\left(x\right)\tan\left(\frac{x}{2}\right)=\frac{cosx}{1+cosx}$
$8b-4b-6$
$4^3-6^2$
$x^2-24x+25=0$
$\left(4x^3+\:2x\right)^2$
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