\(A=x+1+\dfrac{1}{x-1}\\ \\ =x-1+2+\dfrac{1}{x-1}\\ =\left(x-1\right)+\dfrac{1}{x-1}+2\)
Áp dụng \(BDT:\dfrac{a}{b}+\dfrac{b}{a}\ge2\)
\(\Rightarrow A=\left(x-1\right)+\dfrac{1}{x-1}+2\ge2+2\ge4\)
Dấu "=" xảy ra khi:
\(x-1=1\\ \Leftrightarrow x=2\)
Vậy \(A_{Min}=4\) khi \(x=2\)