$\lim_{x\rightarrow 0}\dfrac{(x^2+\pi^{21})\sqrt[7]{1-2x}-\pi^{21}}{x}$
$=\lim_{x\rightarrow 0}\dfrac{x^2.\sqrt[7]{1-2x}+\pi^{21}.\sqrt[7]{1-2x}-\pi^{21}}{x}$
$=\lim_{x\rightarrow 0}x.\sqrt[7]{1-2x}+\lim_{x\rightarrow 0}\pi^{21}.\lim_{x\rightarrow 0}\dfrac{\sqrt[7]{1-2x}-1}{x}$
$A=\lim_{x\rightarrow 0}\dfrac{\sqrt[7]{1-2x}-1}{x}$
Ta có đẳng thức : `x^n-1=(x-1)(x^(n-1)+x^(n-2)+...+x+1)`
$⇒(\sqrt[7]{1-2x})^7-1=(\sqrt[7]{1-2x}-1)((\sqrt[7]{1-2x})^6+(\sqrt[7]{1-2x})^5+...+1)$
$⇒(\sqrt[7]{1-2x})=\dfrac{1-2x-1}{(\sqrt[7]{1-2x})^6+(\sqrt[7]{1-2x})^5+...+1}$
$⇒ \lim_{x\rightarrow 0}\dfrac{\sqrt[7]{1-2x}-1}{x}=\lim_{x\rightarrow 0}\dfrac{-2}{(\sqrt[7]{1-0})^6+(\sqrt[7]{1-0})^5+...+1}=\dfrac{-2}7$
$⇒\lim_{x\rightarrow 0}\dfrac{(x^2+\pi^{21})\sqrt[7]{1-2x}-\pi^{21}}{x}=\dfrac{-2\pi^{21}}{7}$
⇒Chọn A
