Giúp mình câu này với, cảm ơn nhiều.

Đáp án:

$-\dfrac{2\pi^{21}}{7}$

Giải thích các bước giải:

 $\lim_{x\rightarrow 0}\dfrac{x^2\sqrt[7]{1-2x}+\pi ^{21}(\sqrt[7]{1-2x}-1)}{x}$

$=\lim_{x\rightarrow 0}x\sqrt[7]{1-2x}+\lim_{x\rightarrow 0}\dfrac{\pi ^{21}(\sqrt[7]{1-2x}-1)(\sqrt[7]{(1-2x)^6}+\sqrt[7]{(1-2x)^5}+\sqrt[7]{(1-2x)^4}+\sqrt[7]{(1-2x)^3}+\sqrt[7]{(1-2x)^2}+\sqrt[7]{(1-2x)}+1)}{x(\sqrt[7]{(1-2x)^6}+\sqrt[7]{(1-2x)^5}+\sqrt[7]{(1-2x)^4}+\sqrt[7]{(1-2x)^3}+\sqrt[7]{(1-2x)^2}+\sqrt[7]{(1-2x)}+1)}$

$=0+\lim_{x\rightarrow 0}\dfrac{-2\pi ^{21}x}{x(\sqrt[7]{(1-2x)^6}+\sqrt[7]{(1-2x)^5}+\sqrt[7]{(1-2x)^4}+\sqrt[7]{(1-2x)^3}+\sqrt[7]{(1-2x)^2}+\sqrt[7]{(1-2x)}+1)}$

$=\lim_{x\rightarrow 0}\dfrac{-2\pi ^{21}}{\sqrt[7]{(1-2x)^6}+\sqrt[7]{(1-2x)^5}+\sqrt[7]{(1-2x)^4}+\sqrt[7]{(1-2x)^3}+\sqrt[7]{(1-2x)^2}+\sqrt[7]{(1-2x)}+1}$

$=-\dfrac{2\pi^{21}}{7}$