Đáp án + Giải thích các bước giải:
ĐKXĐ : `x \ge 0`
`sqrtx + 2sqrt{x+3} = x + 4`
`⇔ (sqrtx+2sqrt{x+3})^2 = (x+4)^2`
`⇔ x + 4sqrt{x^2+3x} + 4x + 12 = x^2 + 8x + 16`
`⇔ 4sqrt{x^2+3x} = x^2 + 8x + 16 - x - 4x - 12`
`⇔ 4sqrt{x^2+3x} = x^2 + 3x + 4`
`⇔ (4sqrt{x^2+3x})^2 = (x^2+3x+4)^2`
`⇔ 16x^2 + 48 = x^4 + 17x^2 + 16 + 6x^3 + 24x`
`⇔ 16x^2 + 48x - x^4 - 17x^2 - 16 - 6x^3 - 24x = 0`
`⇔ -x^4 - 6x^3 - x^2 + 24x - 16 = 0`
`⇔ -x^4 + x^3 - 7x^3 + 7x^2 - 8x^2 + 8x + 16x - 16 = 0 `
`⇔ -x^3(x-1)-7x^2(x-1)-8x(x-1)+16(x-1) = 0`
`⇔ -(x-1)(x^3+7x^2+8x-16) = 0`
`⇔ -(x-1)(x^3-x^2+8x^2-8x+16x-16) = 0`
`⇔ -(x-1)[x^2(x-1)+8x(x-1)+16(x-1)] = 0`
`⇔ -(x-1)(x-1)(x^2+8x+16) = 0`
`⇔ -(x-1)^2(x+4)^2 = 0`
TH1 : `(x-1)^2 = 0 ⇔ x - 1 = 0 ⇔ x = 1(TM)`
TH2 : `(x+4)^2 = 0 ⇔ x + 4 = 0 ⇔ x = -4(KTM)`
Vậy `S = {1}`