$t=x\left(\frac{kb}{k-b}\right)$
$96.96\cdot16.48$
$\left(2x\right)^3-y^3$
$\frac{d^2}{dx^2}\left(\frac{\left(x-1\right)}{x+1}\right)$
$\left(\frac{y^5}{3x^3}\right)-4$
$\frac{50}{\left(-5\right)}\cdot10$
$2=log4^{16}$
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