$\left(9x+8\right)^3$
$1+\frac{1}{\tan\left(x\right)}$
$12ad-6a+10d-5$
$\frac{26}{8.3}$
$\frac{dy}{dx}=\frac{\left(2xy+3\right)}{x^2}$
$\frac{6x-7}{4}\ge-5$
$4x^2-6x=0$
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