$\frac{\left(4x^2+5x-3\right)}{\left(x+1\right)}$
$x^5-4x^4-21x^3$
$a=21,b=14,\beta\:=100^\circ\:$
$8y\:+\:5z\:-\:6y\:-\:9\:-\:z$
$x^2+4x+6=0$
$\frac{\sqrt{-28}}{10}^2$
$a-5\ge7$
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