$-11\le1-3x$
$x^2\sqrt{x^3\sqrt[4]{x^{-4}}}$
$\frac{x}{5x-6}-\frac{x+3}{6x}$
$y^'-3xy=y^2x$
$\left(-5m+6\right)+\left(-m+5\right)-6$
$\left(y-18\right)\left(y+13\right)$
$4p\:+\:6q\:=\:20$
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