Cho hàm số f(x)= (x - căn(x+ 2)) / (x^2 - 4), x>2 và x^2 + ax + 3b, x<2 và 2a + b - 6, x= 2 liên tục

* Hướng dẫn giải

Đáp án C

Ta có 

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

$=\underset{x\to 2^{+}}{\lim }\frac{x+1}{\left(x+2\right)\left(x+\sqrt{x+2}\right)}=\frac{3}{16}$

Và $\underset{x\to 2^{-}}{\lim }f\left(x\right)=f\left(2\right)=2^2+2a+3b=2a+3b+4$

Do đó $\left\{\begin{array}{l}2a+3b+4=\frac{3}{16}\\ 2a+3b+4=2a+b-6\end{array}\right.\iff \left\{\begin{array}{l}a=\frac{179}{32}\\ b=-5\end{array}\right.$

Vậy $I=a+b=\frac{179}{32}-5=\frac{19}{32}$