2sin x tan x −5 = cos x2sin\:x\:tan\:x\:-5\:=\:cos\:x2sinxtanx−5=cosx
(5−16)−(7−3−6)−(9−13−5)\left(5-16\right)-\left(7-3-6\right)-\left(9-13-5\right)(5−16)−(7−3−6)−(9−13−5)
(h8+1+1)(h8+1−1)\left(\sqrt{h^8+1}+1\right)\left(\sqrt{h^8+1}-1\right)(h8+1+1)(h8+1−1)
x2−4x−5>0x^2-4x-5>0x2−4x−5>0
(12x)2\left(\frac{1}{2x}\right)^2(2x1)2
∫52xx2−25dx\int\frac{5}{2x\sqrt{x^2-25}}dx∫2xx2−255dx
limx→0(ex−1ex−2sinxx2)\lim_{x\to0}\left(\frac{e^x-\frac{1}{e^x}-2\sin x}{x^2}\right)x→0lim(x2ex−ex1−2sinx)
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