(1x−2)5\left(\frac{1}{x}-2\right)^5(x1−2)5
12b⋅2ab12b\cdot2ab12b⋅2ab
m2−2m−15>0m^2-2m-15>0m2−2m−15>0
30x−1230x-1230x−12
(5x2+100)\left(5x^2+100\right)(5x2+100)
limx→∞((πx)(ln(2+ex)))\lim_{x\to\infty}\left(\frac{\left(\pi x\right)}{\left(ln\left(2+e^x\right)\right)}\right)x→∞lim((ln(2+ex))(πx))
x2+1x+4x^2+1x+4x2+1x+4
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