Lời giải:
`A=(x^2-2x+2011)/x^2`
`=1+2/x+2011/x^2`
Đặt `y=1/x` $(y\neq0)$
Khi đó: $A=1+2y+2011y^2$
`=2011(y^2+2.y. 1/2011+1/2011^2)+2010/2011`
`=2011(y+1/2011^2+2010/2011>=2010/2011`
Dấu "=" xảy ra `⇔y+1/2011=0`
`⇔y=-1/2011`
`⇔1/x=-1/2011`
`⇔x=-2011`
Vậy `A_(min)=2010/2011 <=> x=-2011`