`#tnvt`
`\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=0(x>=1)`
`<=>\sqrt{x-1-2.\sqrt{x-1}.2+4}+\sqrt{x-1-2.\sqrt{x-1}.9+9}=0`
`<=>\sqrt{(\sqrt{x-1}-2)^2}+\sqrt{(\sqrt{x-1}-3)^2}=0`
`<=>|\sqrt{x-1]-2|+|\sqrt{x-1}-3|=0`
`∀x>=1` ta có: `{(|\sqrt{x-1}-2|>=0),(|\sqrt{x-1}-3|>=0):}`
`->`Để `|\sqrt{x-1]-2|+|\sqrt{x-1}-3|=0`
Thì `{(|\sqrt{x-1}-2|=0),(|\sqrt{x-1}-3|=0):}`
`<=>{(\sqrt{x-1}-2=0),(\sqrt{x-1}-3=0):}`
`<=>{(\sqrt{x-1}=2),(\sqrt{x-1]=3):}`
`<=>{(x-1=4),(x-1=9):}`
`<=>{(x=5),(x=10):}(\text{Loại})`
Vậy `S={∅}`
