log36m=5\log_{36}m=5log36m=5
limx→−∞(4x4+x2+5x64x8+x6)\lim_{x\to-\infty}\left(\frac{4x^4+x^2+5x}{\sqrt{64x^8+x^6}}\right)x→−∞lim(64x8+x64x4+x2+5x)
limx→∞2tanx5x3\lim_{x\to\infty}\frac{2tanx}{5x^3}x→∞lim5x32tanx
(c−4)(c+9)2\left(c-4\right)\left(c+9\right)^2(c−4)(c+9)2
8b − 5 − 3b8b\:-\:5\:-\:3b8b−5−3b
f(x)=(1x−4x3)(x3−5x−1)f\left(x\right)=\left(\frac{1}{x}-\frac{4}{x^3}\right)\left(x^3-5x-1\right)f(x)=(x1−x34)(x3−5x−1)
−20 12+16.5-20\:\frac{1}{2}+16.5−2021+16.5
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