Với $x=0⇒F=0$
Xét $\dfrac{2002}{F}=\dfrac{2002.(x^2-2x+2002)}{x^2}$
$=\dfrac{2002x^2-4004x+2002^2}{x^2}$
$=2002-2.\dfrac{2002}{x}+(\dfrac{2002}{x})^2$
$=2001+1-2.\dfrac{2002}{x}+(\dfrac{2002}{x})^2$
$=2001+(1-\dfrac{2002}{x})^2$
Mà $(1-\dfrac{2002}{x})^2≥0∀x$
$⇒2001+(1-\dfrac{2002}{x})^2≥2001∀x$
Hay $\dfrac{2002}{F}≥2001∀x$
$⇒F≤\dfrac{2002}{2001}$
Dấu $=$ xảy ra $⇔(1-\dfrac{2002}{x})^2=0⇔1-\dfrac{2002}{x}=0⇔\dfrac{2002}{x}=1⇔x=2002$
Vậy $Max_{F}=\dfrac{2002}{2001}$ tại $x=2002$
