$\frac{\infty^2.47}{\infty^2+47}$
$3x+4\le5x-2$
$3\:x\:3+\left(-9\right)$
$\ \left(2\left(2\right)-5\right)\left(-3\left(2\right)+4\right)\:$
$1.4\cdot10^1$
$\:89mn\:+\:28\:zx\:+\:200zx\:-\:90yk\:+\:70mn-\:25zx+\:45mn\:-\:30yk$
$\frac{x^2-16}{x-5}$
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