ĐK :`1/2 le x le 5`

Đặt `t = sqrt(2x -1)+ sqrt(5- x) (t >0)` ta có:

`t^2 = 2x- 1 + 5 -x + 2sqrt((2x -1)(5- x)) = x +4+ 2sqrt(-2x^2 +11x- 5)`

`-> x -2+ 2sqrt(-2x^2+ 11x -5) = t^2- 6` 

`->t = t^2 -6 -> t^2- t -6 =0 -> (t -3)(t+ 2) =0 -> t= 3` (do `t >0`)

`->x+ 4 + 2sqrt((2x- 1)(5 -x)) =9`

`-> 2sqrt((2x -1)(5- x)) = 5 -x` (`t=3 ->t^2=9`)

`-> 4(2x- 1)(5 -x) = (5- x)^2`

`-> (5 -x)[4(2x -1)- (5 -x)] =0`

`-> (5- x)(9x -9)= 0 `

`->[(x =5),(x= 1):}` (thỏa mãn)

Vậy `x in {1;5}`