$\frac{\left(-6\right)^2}{-6}$
$y'=\frac{x^2}{1+y^2}$
$\lim_{h\to0}\left(\frac{\left(4+h\right)^3-64}{h}\right)$
$2r-r-10r+\:-14=-5$
$\left(-x^2+y^2\right)dx+xydy=0$
$3x^{2-4x-9}$
$\left(1-x\right)\left(-2-x\right)$
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