$ĐKXĐ: x\ne0$
$P=\dfrac{x+1}{x^2}$
$P=\dfrac{x}{x^2}+\dfrac{1}{x^2}$
$P=\dfrac{1}{x}+\dfrac{1}{x^2}$
$P=(\dfrac{1}{x})^2+2.\dfrac{1}{x}.\dfrac{1}{2}+(\dfrac{1}{2})^2-\dfrac{1}{4}$
$P=(\dfrac{1}{x}+\dfrac{1}{2})^2-\dfrac{1}{4}\ge-\dfrac{1}{4}$
Dấu $"="$ xảy ra
$\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{2}=0$
$\Leftrightarrow\dfrac{1}{x}=-\dfrac{1}{2}$
$\Leftrightarrow x=-2$
Vậy $MinP=-\dfrac{1}{4}$ khi $x=-2$
