\(x^2-4x+a=0\) có nghiệm \(x=1\Rightarrow1-4+a=0\Leftrightarrow a=3\)
\(\Rightarrow\lim\limits_{x\rightarrow1}f\left(x\right)=\lim\limits_{x\rightarrow1}\dfrac{x^2-4x+3}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(x-3\right)}{x-1}=-2\)
\(f\left(1\right)=b+2\)
De ham so lien tuc tai x=1\(\Leftrightarrow\lim\limits_{x\rightarrow1}f\left(x\right)=f\left(1\right)\Leftrightarrow b+2=-2\Leftrightarrow b=-4\)
\(\Rightarrow2a+b=2.3-4=2\)
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