$-4x^2-16>0$
$\left(2m^4-6n^5\right)^2$
$3\left(y+5\right)+10\left(y-4\right)+6$
$\lim_{x\to-\infty}\left(\frac{3x+4}{\sqrt[2]{2x^2-5}}\right)$
$4x^4-4x$
$-x^2-6x+27$
$\int\frac{7x^3}{\sqrt{1-4x^2}}dx$
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