Đáp án:
`A=x^2-x+1/x+2013`
`<=>A=x^2-2x+x+1/x+2013`
`<=>A=x^2-2x+1+x-2+1/x+2014`
`<=>A=(x-1)^2+(\sqrt{x}-\sqrt{1/x})^2+2014`
`\text{Vì:}` \(\begin{cases}(x-1)^2 \ge 0\\\left(\sqrt{x}-\sqrt{\dfrac{1}{x}}\right)^2 \ge 0\\\end{cases}\)
`=>(x-1)^2+(\sqrt{x}-\sqrt{1/x})^2 \ge 0`
`<=>(x-1)^2+(\sqrt{x}-\sqrt{1/x})^2+2014 \ge 2014`
Hay `A>=2014`.
Dấu "=" xảy ra khi \(\begin{cases}x-1=0\\\sqrt{x}=\sqrt{\dfrac{1}{x}}\\\end{cases}\)
`<=>` \(\begin{cases}x=1\\x=1\\\end{cases}\)
`<=>x=1(tmđk).`
Vậy \(Min_A=2014\Leftrightarrow x=1\)
