Cho a, b là hai số thực thỏa mãn đi...

A

- Hướng dẫn giải

Phân tích

$$\begin{gathered} \sqrt{4x^2+4x+4}-\left(ax+b\right) \\ =\sqrt{4x^2+4x+1}-\left(2x+1\right) \\ +\left(2x+1\right)-\left(ax+b\right) \\ =\sqrt{4x^2+4x+1}-\left(2x+1\right) \\ +\left(2-a\right)x+1-b \end{gathered}$$

Ta có 

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

Khi đó 

$$\begin{gathered} \underset{x\to \infty }{\lim }\left[\sqrt{4x^2+4x+3}-\left(ax+b\right)\right] \\ \underset{x\to \infty }{\lim }\left[\left(2-a\right)x+1-b\right]=0 \\ \iff \left\{\begin{array}{l}2-a=0\\ 1-b=0\end{array}\right. \\ \iff \left\{\begin{array}{l}a=2\\ b=1\end{array}\right. \end{gathered}$$

Suy ra ${\left(a-2b\right)}^{2018}\left(3a^2+\sqrt[3]{ab}+b\sqrt[3]{a}\right)$=0

Đáp án A