`B=3+\sqrt{-x^2+4x-2}`
ĐKXĐ: `-x^2+4x-2>=0`
`<=> x^2-4x+4-2<=0`
`<=> (x-2)^2-2<=0`
`<=> (x-2-\sqrt{2})(x-2+\sqrt{2})<=0`
`<=>`\(\left[ \begin{array}{l}\begin{cases} x≥2+\sqrt{2}\\x≤2-\sqrt{2}\end{cases}\\\begin{cases} x≤ 2+\sqrt{2}\\x≥2-\sqrt{2}\end{cases}\end{array} \right.\)
`<=> 2-\sqrt{2}<=x<=2+\sqrt{2}`
Do `\sqrt{-x^2+4x-2}>=0`
`=> 3+\sqrt{-x^2+4x-2}>=3`
Dấu = xảy ra khi `-x^2+4x-2=0`
`<=>`\(\left[ \begin{array}{l}x=2+\sqrt{2}(\text{tm})\\x=2-\sqrt{2}(\text{tm})\end{array} \right.\)
Vậy `B_(min)=3<=>x=2+-\sqrt{2}`
