$2x\frac{5}{3}>\frac{x}{3}+10$
$\frac{3x^4-x^2-1}{3x^2-3x-4}$
$\frac{\left(x^2+5\right)^5}{\left(x^3+6\right)^2}$
$b^2+2b-3$
$4x\:-\:8\:-\:8$
$\left(2x^3-11x^2-13x-44\right)\left(x-5\right)$
$\frac{dy}{dx}-4cos\left(4x\right)=-cos\left(4x\right)$
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