$4\left(x^2+2x\right)-3\left(x-5\right)\ge0$
$5\cdot5\cdot5$
$\left(\frac{1}{2}a+3\right)^2$
$7x+3-5x+8+4x-1$
$\frac{\left(-2x^2-6y^2\right)}{y^3-x^2y}$
$\lim_{x\to0}\left(\frac{\sin\left(\pi x\right)}{x^2-4}\right)$
$\frac{6x^2+18x}{2x^3}=\frac{5}{x}$
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