$x^2-5y^2=3$
$3m^2+30m+75$
$-4x+8<-3+x$
$\frac{x^3y^5}{xy^4}$
$\frac{d}{dx}x\sqrt[5]{1+x^4}$
$\lim_{x\to\infty}\left(\frac{ln\left(x-3\right)}{x-3}\right)$
$-3.7\:-\:\left(-4.9\right)$
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