$\lim_{x\to0}\left(\frac{x}{\arctan\left(x\right)^2}\right)$
$-\:6\:+\:7\:-\:8\:-\:5\:+\:12\:-\:17$
$h\left(x\right)=\frac{x^{2}+x-6}{x-2}$
$\frac{1}{2}x^{2}+14x^{5}-1^{2}+3x-6y^{3}\right)$
$-1+\left(25-1\right)\left(-10\right)$
$\frac{x+8}{6-x}=13$
$-7\cdot x<7$
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