Tính lim(x->1) ( căn bậc 3 (x+7) - căn bậc 2(x^2 +x+2))/(x-1)

* Hướng dẫn giải

Đáp án D

$$\begin{gathered} \underset{x\to 1}{\lim }\frac{\sqrt[3]{x+7}-\sqrt{x^2+x+2}}{x-1} \\ =\underset{x\to 1}{\lim }\frac{\sqrt[3]{x+7}-2}{x-1}-\underset{x\to 1}{\lim }\frac{\sqrt[]{x^2+x+2}-2}{x-1} \end{gathered}$$

$$\begin{gathered} =\underset{x\to 1}{\lim }\frac{1}{\sqrt[3]{{\left(x+7\right)}^2}+2\sqrt[3]{\left(x+7\right)}+4}-\underset{x\to 1}{\lim }\frac{x+2}{\sqrt[]{x^2+x+2}+2} \\ =\frac{1}{12}-\frac{3}{4}=-\frac{2}{3} \end{gathered}$$