$3x\:+\:5\:>\:2x\:-8$
$\left(3y+5\right)^2$
$\left(\frac{1}{4}+ab\right)^2$
$xy'+y=\left(x+1\right)^2$
$xy^2dx+e^xdy=0$
$8+4+6+9+19+12+15-19$
$\frac{1}{81}x^2-4>0$
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