$-2=v-4$
$g\left(x\right)=tan\left(3-4\right)+2$
$\frac{7}{x+2}+\frac{5}{x-2}=\frac{10x-2}{x^2-4}$
$\lim_{x\to\infty}\left(\frac{ln\left(3x+8\right)}{ln\left(2x+3\right)+4}\right)$
$9b^7-4b^8+3b^5$
$\left(x^{2}+5x-4\right)\left(2x^{3}+3x-1$
$\frac{\infty}{-\infty+\infty+6}$
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