ĐK : $\begin{cases} 3x-3 ≥ 0\\5 -x≥0\\2x-4≥0 \end{cases}$ `<=>` $\begin{cases} x≥1\\x≤5\\x≥2 \end{cases}$
`=>` `2 ≤ x ≤ 5`
`sqrt(3x-3) - sqrt(5-x) = sqrt(2x-4)`
`<=> 3x-3 - 2sqrt((3x-3)(5-x)) + 5-x = 2x-4`
`<=> 2x + 2 - 2sqrt((3x-3)(5-x)) = 2x-4`
`<=> 6 - 2sqrt((3x-3)(5-x)) = 0`
`<=> 2sqrt((3x-3)(5-x)) = 6`
`<=> sqrt((3x-3)(5-x)) = 3`
`<=> (3x-3)(5-x) = 9`
`<=> 15x - 3x^2 - 15 + 3x = 9`
`<=> x^2 - 6x + 8 = 0`
`<=> (x-2)(x-4) = 0`
`->` $\left[\begin{matrix} x-2=0\\ x-4=0\end{matrix}\right.$
`->`$\left[\begin{matrix} x=2\\ x=4\end{matrix}\right.$
Vậy phương trình có `2` nghiệm : `x = 2 ; x = 4`
