Question

Answer 1

`A=sqrt(1-2x+x^2)+sqrt(x^2 -6x+9)`

`=sqrt((1-x)^2)+sqrt((x-3)^2)`

`=|1-x|+|x-3|`

`>= |1-x+x-3|=|-2|=2`

Dấu "=" xảy ra `<=>(1-x)(x-3) >= 0 <=> 1 <= x <= 3`

Vậy `A_(min) =2 <=> 1 <= x <= 3`

Answer 2

\(A=\sqrt{\left(x-1\right)^2}+\sqrt{\left(3-x\right)^2}=\left|x-1\right|+\left|3-x\right|\ge\left|x-1+3-x\right|=2\)

\(A_{min}=2\) khi \(1\le x\le3\)