$10x^8y^6:2x^3y^4$
$\lim_{x\to-1}\left(x+4\right)$
$\left(9x^2+7x-6\right)-\left(3x^2-8x+7\right)$
$724.7+329+523.25$
$\lim_{x\to-\infty}\frac{-5}{2x^3}-7+\frac{8}{x}$
$x^4+7x^2+15x=7x^3$
$12a^6-75b^8$
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