$\lim_{b\to0}\sqrt{1+b^3}$
$\frac{s^{-8}t^0u^{-9}\cdot s^0tu^{-7}}{s^{-4}tu^8}$
$\frac{1}{t-4}=\frac{t}{t+24}$
$\left(xyz+xy^2z^2\right)^2$
$2x+y=-14$
$x\cdot\tan\left(47\right)=\left(x+389\right)\tan\left(17\right)$
$x^2+6x+28$
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