$$\begin{gathered} \underset{x\to 0}{\lim }\frac{\left(x^2+2012\right)\sqrt[7]{1-2x}-2012}{x}=\underset{x\to 0}{\lim }\left(x\sqrt[7]{1-2x}\right)+2012\underset{x\to 0}{\lim }\frac{(\sqrt[7]{1-2x}-1)}{x} \\ =2012\underset{x\to 0}{\lim }\frac{\sqrt[7]{1-2x}-1}{x} \end{gathered}$$

Xét hàm số $\text{y}=\text{f}\left(\text{x}\right)=\sqrt[7]{1-2\text{x}}$ ta có f(0) = 1. Theo định nghĩa đạo hàm ta có:

$$\begin{gathered} f^{\text{'}}\left(0\right)=\underset{x\to 0}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x-0}=\underset{x\to 0}{\lim }\frac{\sqrt[7]{1-2x}-1}{x} \\ {f}^{\text{'}}\left(x\right)=-\frac{2}{7{\left(\sqrt[7]{1-2x}\right)}^6}\Rightarrow f\left(0\right)=-\frac{2}{7}\Rightarrow \underset{x\to 0}{\lim }\frac{\sqrt[7]{1-2x}-1}{x}=-\frac{2}{7} \\ \Rightarrow \underset{x\to 0}{\lim }\frac{\left(x^2+2012\right)\sqrt[7]{1-2x}-2012}{x}=-\frac{4024}{7} \\ \Rightarrow \left\{\begin{array}{l}a=-4024\\ b=7\end{array}\right.\Rightarrow a+b=-4017 \end{gathered}$$