Đặt f(x) = \(x^2+ax+b\Rightarrow f\left(2\right)=2a+b+4\)
Ta có : \(lim_{x\rightarrow2}\dfrac{f\left(x\right)}{x-2}=6\) \(\Rightarrow f\left(2\right)=0\) \(\Rightarrow2a+b+4=0\) \(\Leftrightarrow a=\dfrac{-b-4}{2}\)
Khi đó : \(f\left(x\right)=x^2-\dfrac{b+4}{2}x+b=\left(x-2\right)\left(x-\dfrac{b}{2}\right)\)
\(Lim_{x\rightarrow2}\dfrac{f\left(x\right)}{x-2}=Lim_{x\rightarrow2}\left(x-\dfrac{b}{2}\right)=2-\dfrac{b}{2}=6\)
\(\Rightarrow b=-8\Rightarrow a=2\) => a + b = -6
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