$\frac{6x}{\left(x^2+4x-5\right)}$
$-4\frac{3}{4}$
$10\:+\:b\:\le\:-1$
$144-4x^3+8x^2$
$cos\:t\:=-\frac{4}{5}$
$\left(-3\right)-\left(-2\right)+\left(-6\right)\cdot\left(-15\right)+18$
$\lim_{x\to2}\left(\frac{\sqrt[3]{5x-2}-\sqrt[3]{x+6}}{x^3-4}\right)$
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