ĐK: \(x\ge1\)
\(\sqrt{x-2\sqrt{x-1}}=\sqrt{x-1-2\sqrt{x-1}+1}=\sqrt{\left(\sqrt{x-1}-1\right)^2}=\left|\sqrt{x-1}-1\right|\)
Tương tự \(\sqrt{x+3-4\sqrt{x-1}}=\sqrt{\left(\sqrt{x-1}-2\right)^2}=\left|\sqrt{x-1}-2\right|\)
\(\sqrt{x+8-6\sqrt{x-1}}=\sqrt{\left(\sqrt{x-1}-3\right)^2}=\left|\sqrt{x-1}-3\right|\)
\(\Rightarrow A=\left|\sqrt{x-1}-1\right|+5\left|\sqrt{x-1}-2\right|+\left|\sqrt{x-1}-3\right|\)
\(\Rightarrow A=\left|\sqrt{x-1}-1\right|+\left|3-\sqrt{x-1}\right|+5\left|\sqrt{x-1}-2\right|\)
\(\Rightarrow A\ge\left|\sqrt{x-1}-1+3-\sqrt{x-1}\right|+5\left|\sqrt{x-1}-2\right|=5\left|\sqrt{x-1}-2\right|+2\ge2\)
\(\Rightarrow A_{min}=2\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\sqrt{x-1}-1\ge0\\\sqrt{x-1}-3\le0\\\sqrt{x-1}-2=0\end{matrix}\right.\) \(\Rightarrow\sqrt{x-1}=2\Rightarrow x=5\)
