$c=\sec x\cdot\cot x+\cos x$
$64x^3\:+\:16x^2\:+\:x$
$\left(a^2b+2\right)\left(a^2b-10\right)$
$0\cdot\left(-10\right)\cdot\left(-19\right)\cdot\left(-38\right)$
$3x+6>0$
$\lim_{x\to\infty}\left(\frac{x+4}{x}\right)^x$
$\frac{12x^2-6x+2}{3}$
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