Đáp án:
`A=\sqrt{x^2-2x+1}+\sqrt{x^2-6x+9}`
`<=>A=\sqrt{(x-1)^2}+\sqrt{(x-3)^2}`
`<=>A=|x-1|+|x-3|`
Vì `|x-1|>=x-1`
`|x-3|>=3-x`
`=>A>=x-1+3-x=2`
Dấu "=" xảy ra khi \(\begin{cases}x-1 \ge 0\\x-3 \le 0\\\end{cases}\)
`<=>` \(\begin{cases}x \ge 1\\x \le 3\\\end{cases}\)
`<=>1<=x<=3`.