Tính lim x đến 1 cưn 3 26 +x - căn x+8/ x^2-3x+2

Đáp án đúng là: A

$$\begin{gathered} \underset{x\to 1}{\lim }\frac{\sqrt[3]{26+x}-\sqrt{x+8}}{x^2-3x+2} \\ =\underset{x\to 1}{\lim }\frac{\left(\sqrt[3]{26+x}-3\right)+\left(3-\sqrt{x+8}\right)}{x^2-3x+2} \\ =\underset{x\to 1}{\lim }\left[\frac{\left(\sqrt[3]{26+x}-3\right)\left({\sqrt[3]{26+x}}^2+\sqrt[3]{26+x}+9\right)}{\left(x-1\right)\left(x-2\right)\left({\sqrt[3]{26+x}}^2+3\sqrt[3]{26+x}+9\right)}+\frac{\left(3-\sqrt{x+8}\right)\left(3+\sqrt{x+8}\right)}{\left(x-1\right)\left(x-2\right)\left(3+\sqrt{x+8}\right)}\right] \\ =\underset{x\to 1}{\lim }\left[\frac{\left(26+x\right)-27}{\left(x-1\right)\left(x-2\right)\left({\sqrt[3]{26+x}}^2+3\sqrt[3]{26+x}+9\right)}+\frac{9-\left(x+8\right)}{\left(x-1\right)\left(x-2\right)\left(3+\sqrt{x+8}\right)}\right] \\ =\underset{x\to 1}{\lim }\left[\frac{1}{\left(x-2\right)\left({\sqrt[3]{26+x}}^2+3\sqrt[3]{26+x}+9\right)}-\frac{1}{\left(x-2\right)\left(3+\sqrt{x+8}\right)}\right] \\ =-\frac{1}{27}+\frac{1}{6}=\frac{7}{54}. \end{gathered}$$