`A=\sqrt{x^2+3x-4}+8`
ĐKXĐ: `x^2+3x-4>=0`
`<=> x^2+4x-x-4>=0`
`<=> x(x+4)-(x+4)>=0`
`<=> (x+4)(x-1)>=0`
`<=>`\(\left[ \begin{array}{l}x≤-4\\x≥1\end{array} \right.\)
Do `\sqrt{x^2+3x-4}>=0` với `AAx`
`-> \sqrt{x^2+3x-4}+8>=8`
`-> A_(min)=8`
Dấu = xảy ra khi `\sqrt{x^2+3x-4}=0`
`<=>`\(\left[ \begin{array}{l}x=-4\\x=1\end{array} \right.\)
Vậy `A_(min)=8<=> x∈{-4;1}`
