limx→1(x2−8x+7x2−2x+1)\lim_{x\to1}\left(\frac{x^2-8x+7}{x^2-2x+1}\right)x→1lim(x2−2x+1x2−8x+7)
∫x(x2+x+1)dx\int x\left(x^2+x+1\right)dx∫x(x2+x+1)dx
2v−3=112v-3=112v−3=11
6(7x+1)6\left(7x+1\right)6(7x+1)
∞=∞\infty=\infty∞=∞
(u+1)2\left(u+1\right)^2(u+1)2
−(−3(a−b)+13(−6a−9b))-\left(-3\left(a-b\right)+\frac{1}{3}\left(-6a-9b\right)\right)−(−3(a−b)+31(−6a−9b))
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