$\sqrt(x)+$\sqrt(2x+1)=2

ĐKXĐ: `x \ge 0`

`sqrtx + sqrt(2x+1) = 2`

Bình phương `2` vế:

`=> x + 2sqrt(x(2x+1)) + 2x + 1 = 4`

`<=> x+2sqrt(2x^2+x) + 2x + 1 = 4`

`<=> 3x + 2sqrt(2x^2+x) + 1 = 4`

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

Bình phương `2` vế:

`=> 4(2x^2+x) = 9 - 18x + 9x^2`

`<=> 8x^2 + 4x = 9 - 18x + 9x^2`

`<=> 8x^2 + 4x - 9 + 18x - 9x^2 = 0`

`<=> x^2 - 22x + 9 = 0`

`<=> x^2 - 22x + 121 = -9 + 121`

`<=> (x-11)^2 = 112`

`@ x - 11 = 4sqrt7`

`<=> x = 4sqrt7 + 11` (TM)

`@ x - 11 = -4sqrt7`

`<=> x = -4sqrt7 + 11` (L)

Vậy `S = {4sqrt7+11}`