$+\frac{3}{4}xy^2+\frac{1}{8}xy^2$
$2=\log\left(2\right)+\log\left(x\right)$
$\left(\frac{1}{x-2}\right)-\left(\frac{6}{x^2+2x-8}\right)$
$2\left(-14+r\right)-\left(-3r-0\right)$
$x^8-16x^4+64$
$\frac{x^3+3}{x-1}$
$\lim_{x\to1}\left(\frac{4x^2lnx}{x^2-1}\right)$
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