\(A=\dfrac{x^2+x+1}{x}\\ \\ =\dfrac{x^2}{x}+\dfrac{x}{x}+\dfrac{1}{x}\\ \\ =x+1+\dfrac{1}{x}\)
Áp dụng BDT: Cô-si: \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\)
\(\Rightarrow A=x+1+\dfrac{1}{x}\ge2+1\ge3\)
Dấu "=" xảy ra khi:
\(x=1\)
Vậy \(A_{\left(Min\right)}=3\) khi \(x=1\)