Cho các đa thức f(x), g(x) thỏa mãn $\underset{x\to 1}{\lim }\frac{f\left(x\right)-4}{x-1}=3$ và $\underset{x\to 1}{\lim }\frac{g\left(x\right)-2}{x-1}=2$.

Biết $\underset{\text{x}\to 1}{\lim }\frac{\sqrt{\text{f}\left(\text{x}\right)\text{g}\left(\text{x}\right)+1}-3}{\text{x}-1}=\frac{\text{a}}{\text{b}}$, vói $\frac{\text{a}}{\text{b}}$ là phân số tối giản. Tính a + b.

Ta có $\left\{\begin{array}{l}\underset{x\to 1}{\lim }\frac{f\left(x\right)-4}{x-1}=3\Rightarrow \underset{x\to 1}{\lim }f\left(x\right)=4\\ \underset{x\to 1}{\lim }\frac{g\left(x\right)-2}{x-1}=2\Rightarrow \underset{x\to 1}{\lim }g\left(x\right)=2\end{array}\right.$

$$\begin{gathered} \Rightarrow L=\underset{x\to 1}{\lim }\frac{\sqrt{f\left(x\right)g\left(x\right)+1}-3}{x-1}=\underset{x\to 1}{\lim }\frac{f\left(x\right)g\left(x\right)-8}{(x-1)[\sqrt{f\left(x\right)g\left(x\right)+1}+3]}=\underset{x\to 1}{\lim }\frac{\left[f\right(x)-4]g\left(x\right)+4\left[g\right(x)-2]}{(x-1)[\sqrt{f\left(x\right)g\left(x\right)+1}+3]} \\ =\underset{x\to 1}{\lim }\frac{f\left(x\right)-4}{x-1}\cdot \frac{g\left(x\right)}{\sqrt{f\left(x\right)g\left(x\right)+1}+3}+4\underset{x\to 1}{\lim }\frac{g\left(x\right)-2}{x-1}\cdot \frac{1}{\sqrt{f\left(x\right)g\left(x\right)+1}+3} \\ =3.\frac{2}{\sqrt{4\cdot 2+1}+3}+4.\frac{2}{\sqrt{4\cdot 2+1}+3}=1+\frac{4}{3}=\frac{7}{3}=\frac{a}{b}\Rightarrow a+b=10 \end{gathered}$$