\(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=5\)
ĐKXĐ:.....
\(\Leftrightarrow\sqrt{x-1-4\sqrt{x-1}+4}+\sqrt{x-1-6\sqrt{x-1}+9}=5\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}-2\right)^2}+\sqrt{\left(\sqrt{x-1}-3\right)^2}=5\)
\(\Leftrightarrow\left|\sqrt{x-1}-2\right|+\left|\sqrt{x-1}-3\right|=5\)
Đặt t = \(\sqrt{x-1}\) ( t ≥ 0)
Ta có pt mới : \(\left|t-2\right|+\left|t-3\right|=5\left(1\right)\)
Xét t - 2 ≥ 0 ⇔ t ≥ 2
t - 3 ≥ 0 ⇔ t ≥ 3
Bảng xét dấu :
| t | 2 3 |
| t - 2 | - 0 + + + |
| t - 3 | - - - 0 + |
Xét TH1 : t ≤ 2
( 1 ) ⇔ 2 - t +3 - t = 5
⇔ -2t = 0
⇔ t = 0 ( N )
TH2 : 2 < t < 3
( 1 ) ⇔ t - 2 +3 - t = 5
⇔ 1 = 5 (VL)
TH3 : t ≥ 3
(1) ⇔t - 2 + t -3 =5
⇔ 2t = 10
⇔ t = 5 ( N )
Vậy 2 ≤ t ≤ 3
Do đó: 2 ≤ \(\sqrt{x-1}\) ≤ 3
\(\left\{{}\begin{matrix}\text{}\sqrt{x-1}\ge2\\\sqrt{x-1}\le3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1\ge4\\x-1\le9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge5\\x\le10\end{matrix}\right.\)
Vậy 5 ≤ x ≤ 10
