$\frac{-8a^2x^3}{-8ax}$
$\frac{d}{dx}9x^2+3x+5$
$\tan^2\left(a\right)+\sin^2\left(a\right)=\left(\sec\left(a\right)+\cos\left(a\right)\right)\left(\sec\left(a\right)-\cos\left(a\right)\right)$
$\left(-10x+y^2+2z\right)^2$
$\left(2x+3y^3z^2\right)^2$
$5<9-x$
$964\cdot8$
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