$\left(\frac{4}{5}x^2\right)\left(x^3-2x^2+3x+2\right)$
$\frac{2a^4b^2}{7a^3b^2}$
$\frac{1}{4}x-3\le\frac{1}{8}\left(3+2x\right)$
$\sin\left(x\right)\tan\left(ydx\right)-\frac{dy}{\sin\left(x\right)}=0$
$36x^2\:-\:196$
$\frac{1}{s^2}-\frac{8}{s}+\frac{1}{s-1}$
$3x>5$
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