limx→∞(x)1ln(x)\lim_{x\to\infty}\left(x\right)^{\frac{1}{ln\left(x\right)}}x→∞lim(x)ln(x)1
∫x(6x2+3x−4)dx\int x\left(6x^2+3x-4\right)dx∫x(6x2+3x−4)dx
limx→−2(4x2+7x−2x2−5x+14)\lim_{x\to-2}\left(\frac{4x^2+7x-2}{x^2-5x+14}\right)x→−2lim(x2−5x+144x2+7x−2)
cos=−25cos=-\frac{2}{5}cos=−52
x2+y4\frac{x}{2}+\frac{y}{4}2x+4y
(x5+32)(x+2)\left(x^5+32\right)\left(x+2\right)(x5+32)(x+2)
∫5∞(1x2−5x+4)dx\int_5^{\infty}\left(\frac{1}{x^2-5x+4}\right)dx∫5∞(x2−5x+41)dx
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