\(P=\dfrac{x+1}{x^2}\left(x\ne0\right)\)
\(=\dfrac{1}{x}+\dfrac{1}{x^2}\)
\(=\left[\left(\dfrac{1}{x}\right)^2+2\cdot\dfrac{1}{x}\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2\right]-\left(\dfrac{1}{2}\right)^2\)
\(=\left(\dfrac{1}{x}+\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
Vậy: \(P_{Min}=-\dfrac{1}{4}\Leftrightarrow x=-2\)