$y^2-6y>-10$
$x^2+\frac{1}{3}+\frac{1}{36}$
$4x^8-24x+36$
$512+8m^6+\left(-8m^2-32\right)\left(m^4-4m^2+16\right)=0$
$\lim_{x\to\infty}\left(\frac{tan\left(x\right)}{5x}\right)$
$\int\left(2x-3\right)^2dx$
$\left(2x^2-3x+6\right)\left(3x^4-5x^3-6x^2-2x-1\right)$
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