$\sqrt{w^{12}x^5z^8}$
$2x+3x-3x$
$\int\left(x+4\right)e^2dx$
$\frac{\cot\left(-t\right)+\tan\left(-t\right)}{\sin\left(t\right)}=-\sec\left(t\right)$
$\left(\sqrt{10c}\right)^2$
$\left(17-5i\right)-\left(16-11i\right)$
$\left(4x+xy^2\right)dx+\left(y^2+x^2y\right)dy=0$
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