Đặt \(A=\sqrt{x+8}+\sqrt{x-1}\)
Xét \(A^2=\left|x+8\right|+2\sqrt{\left(x+8\right)\left(x-1\right)}+\left|x-1\right|\)
\(\Leftrightarrow A^2=x+8+1-x+2\sqrt{\left(x+8\right)\left(x-1\right)}\)
\(\Leftrightarrow A^2=9+2\sqrt{\left(x+8\right)\left(x-1\right)}\)
Ta có \(A^2\ge9\)\(\Rightarrow\)min A = 3 \(\Leftrightarrow x=1\)