$\frac{x^2-8x+3}{x-5}$
$-\frac{53}{8}+\frac{40x^4+90x^2-93x+289}{8\left(2x^2-x+5\right)}$
$-\left(-5-1\right)$
$\lim_{x\to\infty}\left(\frac{3x^5+5}{x^5-x^3+x+9}\right)$
$-34c+c$
$y=\tan\left(8x^2-x\right)$
$\frac{x^2-16}{4-x}=0$
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