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$x^2+x-24=0$
$\frac{1}{\left(1+\frac{z}{r}\right)^2}$
$\left(y+12\right)^2=9\left(x-9\right)$
$\frac{7}{3}\ge x$
$6x^2+6x+3$
$+4+5-8+2-9+4-8+\left(-6\right)$
$x^3\:+\:y^3\:=\:\left(x\:+\:y\right)\left(x^2\:-\:xy\:+\:y^2\right)$
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