Giả thiết suy ra \(f\left(1\right)-2=0\Rightarrow f\left(1\right)=2\)
a.
\(\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-2-x+1}{x^2-3x+2}=\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-2}{x^2-4x+3}.\dfrac{x-3}{x-2}-\lim\limits_{x\rightarrow1}\dfrac{x-1}{\left(x-1\right)\left(x-2\right)}\)
\(=2.\dfrac{1-3}{1-2}-\dfrac{1}{1-2}\)
b.
\(\lim\limits_{x\rightarrow1}\dfrac{\left[f\left(x\right)-1\right]\left[f\left(x\right)-2\right]}{\left(x-1\right)\left(x-2\right).\left[f\left(x\right)+\sqrt{f\left(x\right)+2}\right]}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-2}{x^2-4x+3}.\dfrac{x-3}{x-2}.\dfrac{1}{f\left(x\right)+\sqrt{f\left(x\right)+2}}\)
\(=2.\dfrac{1-3}{1-2}.\dfrac{1}{2+\sqrt{2+2}}\)
\(2=\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-2}{x^2-4x+3}=\lim\limits_{x\rightarrow1}\dfrac{f'\left(x\right)}{2x-4}=\dfrac{f'\left(1\right)}{-2}\)
\(\Rightarrow f'\left(1\right)=-4\)
Đồng thời \(f\left(1\right)-2=0\Rightarrow f\left(1\right)=2\)
a.
\(\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-x-1}{x^2-3x+2}=\lim\limits_{x\rightarrow1}\dfrac{f'\left(x\right)-1}{2x-3}=\dfrac{f'\left(1\right)-1}{2.1-3}=\dfrac{-4-1}{-1}=5\)
b.
\(\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-\sqrt{f\left(x\right)+2}}{x^2-3x+2}=\lim\limits_{x\rightarrow1}\dfrac{f'\left(x\right)-\dfrac{f'\left(x\right)}{2\sqrt{f\left(x\right)+2}}}{2x-3}=\dfrac{-4-\dfrac{-4}{2\sqrt{2+2}}}{2.1-3}\)
