Bài `6:`

`a) sqrt{4x - 4} =16`

ĐK : `x \ge 1`

`<=> 4x - 4 = 256`

`<=> 4x = 256 + 4`

`<=> 4x =260`

`<=> x = 65`

Vậy `S = {65}`

`b) sqrt{x} . (3sqrt{x} + 5) = 3x - sqrt{x} + 4`

ĐK : `x \ge 0`

`to 3x + 5sqrt{x} = 3x - sqrt{x} + 4`

`to 3x + 5sqrt{x} - 3x + sqrt{x} = 4`

`to 6sqrt{x} = 4`

`to sqrt{x} = 2/3`

`to x = 4/9` (tm)

Vậy `S={4/9}`

`c) sqrt{2x+3} - sqrt{18x+27} = -4`

ĐK : `x \ge -3/2`

`<=> sqrt{2x + 3} - sqrt{ 9(2x + 3)} = -4`

`<=> sqrt{2x + 3} - 3sqrt{2x + 3} = -4`

`<=> -2sqrt{2x + 3} = -4`

`<=> sqrt{2x+3} = 2`

`<=> 2x + 3 = 4`

`<=> 2x = 1`

`<=> x = 1/2` (tm)

Vậy `S = {1/2}`

`d) 5/2 . sqrt{4x-16} - sqrt{ (9x-36)/25} = 22`

ĐK : `x \ge 4`

`<=> 5/2 . sqrt{ 4(x - 4)} - sqrt{ 9/25 . (x -4)} = 22`

`<=> 5/2 . 2sqrt{x - 4} - 3/5 sqrt{x -4} = 22`

`<=> 5sqrt{x - 4} - 3/5 sqrt{x - 4} = 22`

`<=> 22/5 . sqrt{x - 4} = 22`

`<=> sqrt{x - 4} = 5`

`<=> x - 4 = 25`

`<=> x = 29` (tm)

Vậy `S = {29}`

`e) sqrt{x^2 + 4x + 4} = sqrt{9x^2 - 36}`

ĐK : `x \ge 2 ; x \le -2`

`<=> sqrt{(x+2)^2} - sqrt{9(x^2 - 4)} = 0`

`<=> sqrt{(x+2)^2} - 3sqrt{ (x + 2)(x - 2)} = 0`

`<=> sqrt{x+2} . (sqrt{x + 2} - 3sqrt{x - 2}) = 0`

`<=> sqrt{x + 2} = 0` hoặc `sqrt{x + 2} - 3sqrt{x - 2} = 0`

`<=> x + 2 = 0` hoặc `sqrt{x + 2} = 3sqrt{x - 2}`

`<=> x = -2` (tm) hoặc `x + 2 = 9(x - 2)`

`<=> x + 2 = 9x - 18` hoặc `x = -2`

`<=> 8x = 20` hoặc `x = -2`

`<=> x = 5/2` (tm) hoặc `x = -2`

Vậy `S = {5/2 ; -2}`

`f) sqrt{x^2 - 6x + 9} + sqrt{x^2 - 9} = 0`

ĐK : `x \ge 3 ; x \le -3`

Ta có : `sqrt{x^2 - 6x + 9} \ge 0 AA x \ge 3 ; x \le -3`

`sqrt{x^2 - 9} \ge 0 AA x \ge 3 ; x \le -3`

`to sqrt{x^2 - 6x + 9} + sqrt{x^2 - 9} \ge 0 AA x \ge 3 ; x \le -3`

Dấu "=" xảy ra khi : `{(x^2 - 6x + 9 = 0),(x^2 - 9 = 0):}`

`<=> {((x-3)^2 = 0),((x-3)(x+3)=0):}`

`<=> x = 3` (tm)

Vậy `S = {3}`