$\lim\:_{x\to\:\infty\:}\left(\frac{3x+4}{3x-2}\right)^{2x+1}$
$2n-3\sqrt{6n^2-7n-5}$
$6x^4-35x^3+62x^2-35x+6$
$\frac{x^5}{5}\left(5\ln x+1\right)$
$+8\left(x+3\right)=m\cdotn$
$\left(y^7+8y\right)\left(y^7-8y\right)$
$x^2-x+8\ge2x^2-2$
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