\(f\left(x\right)=x+\dfrac{1}{x-1}=x-1+\dfrac{1}{x-1}+1\ge2\sqrt{\left(x-1\right).\dfrac{1}{x-1}}+1=3\)
\(\Leftrightarrow f\left(x\right)\ge3\). \("="\Leftrightarrow x-1=\dfrac{1}{x-1}\Leftrightarrow\left(x-1\right)^2=1\Leftrightarrow x=2\left(x>1\right)\)
Đáp án A