Bài 9:

`sqrt(x^2 - 6sqrt(x^2) + 9) + sqrt(x^2) - 7 = 0`

`sqrt( sqrt(x^2)^2 - 2  * 3 * sqrt(x^2) + 3^2)  + |x| -7 =0`

`sqrt( (sqrt(x^2) - 3 )^2) + |x| -7 =0`

`||x| -3| + |x| =7`

TH1:

`0<=|x| < 3`

`=> 3 - |x| + |x| = 7`

`=> 3 = 7` ( vô lí )

TH2"

`|x| >=3`

`=> |x|-3 + |x| = 7`

`=> 2|x| = 10`

`=> x = +-5`

_____________

b)

`sqrt(2x + 4 - 6sqrt(2x -5)) + sqrt(2x - 4 + 2sqrt(2x -5)) = 4`

ĐK: `2x - 5 >= 0`

`x>= 2,5`

Ta có:

+) `sqrt(2x + 4 - 6sqrt(2x -5))`

`= sqrt(sqrt((2x -5)^2) - 2*sqrt(2x -5) *3 + 3^2)`

`= sqrt( (sqrt(2x -5) - 3 )^2 )`

`= |sqrt(2x - 5) - 3|`

+) `sqrt(2x - 4 + 2sqrt(2x -5))`

`= sqrt(sqrt((2x -5)^2) + 2*sqrt(2x-5) *1 + 1^2)`

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

`= sqrt(2x - 5) + 1`

`=> |sqrt(2x - 5) - 3| + sqrt(2x - 5) + 1 = 4`

`=> |sqrt(2x-5) -3| + sqrt(2x-5) = 3`

TH1:

`sqrt(2x - 5) >=3`

`=> sqrt(2x-5) -3 + sqrt(2x-5) = 3`

`=> 2sqrt(2x-5) = 6`

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

TH2:

`0<= sqrt(2x-5) < 3` (2)

`=> 3 - sqrt(2x-5) + sqrt(2x-5) = 3` ( luôn đúng ) (3)

Từ (1) (2) (3) suy ra:

`0 <= sqrt(2x-5) <= 3`

`0 <= 2x - 5 <=9`

`5 <= 2x <=14`

`2,5 <=x<=7`

Vậy...