$x\sqrt{1+y^2}\:+y\sqrt{1+x^2}\frac{dy}{dx}=0$
$\left(10c\:+\:6\right)\left(10c\:-\:\:6\right)$
$\left(-\sqrt{49}+3^4:3^2\right)^5$
$x^2-4x\ge0$
$-\frac{3}{5}m+\frac{1}{4}-m$
$-2a-3a-a^4-a^4+3a^8$
$\int\frac{1+3x}{x^2+4x+6}dx$
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