$2\left(q-3\right)$
$\frac{2x^3-x^2+3x+4}{2x+1}$
$\frac{8z^6-y^9}{2z^2-x}$
$\:\frac{\left(-4\right)^5}{\left(-4\right)^2}$
$\sec\left(x\right)-\sec\left(x\right)\cdot\tan\left(x\right)$
$+4-22-18-8-25$
$\left(23y^2\right)\left(3x\right)$
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