\(\sqrt{x+3-4\sqrt{x-1}}=\sqrt{x-1-4\sqrt{x-1}+4}=\left(\sqrt{x-1}-2\right)^2\)
Và \(\sqrt{x+8+6\sqrt{x-1}}=\sqrt{x-1+6\sqrt{x-1}+9}=\left(\sqrt{x-1}-3\right)^2\)
Ok dễ nhé
ĐKXĐ: \(x\ge1\)
Ta có: \(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8+6\sqrt{x-1}}\)
\(=\sqrt{4-2.2.\sqrt{x-1}+x-1}+\sqrt{x-1+2.\sqrt{x-1}.3+9}\)
\(=\sqrt{\left(2-\sqrt{x-1}\right)^2}+\sqrt{\left(\sqrt{x-1}+3\right)^2}\)\(=|2-\sqrt{x-1}|+|\sqrt{x-1}+3|\ge|2-\sqrt{x-1}+\sqrt{x-1}+3|=5\)
Dấu bằng xảy ra khi \(2-\sqrt{x-1}\ge0\Leftrightarrow\sqrt{x-1}\le2\Leftrightarrow x\le3\)
Vậy \(1\le x\le3\)
Nếu đúng cho nhé bạn.
