Question

\(x^2+2x+7=3\sqrt{\left(x^2+1\right)\left(x+3\right)}\)

Answer 1

Đk: `x >= -3`.

Đặt `{(sqrt(x^2+1)=a), (sqrt(x+3)=b):}`

PT trở thành: `a - 3ab + 2b = 0`

`<=> a(a-b) - 2b(a-b) = 0`

`<=> (a-2b)(a-b) = 0`

`@ a= 2b -> sqrt(x^2+1) = 2 sqrt(x+3)`

`<=> x^2 + 1 = 4x + 12`

`<=> x^2 - 4x - 11  =0`

`\Delta = 16 + 44 = 60 > 0`

`-> x = (4 +-sqrt 60)/2`.

`@ a = b -> sqrt(x^2+1) = sqrt(x+3)`

`<=> x^2 - x - 2 = 0`

`<=> (x-2)(x+1) = 0`

`<=> x=  2` hoặc `-1`.

Vậy PT có ...

Answer 2

ĐKXĐ $ x >= - 3$

$ PT <=> (x^{2} + 1) - 3\sqrt{(x^{2} + 1)(x + 3)} + 2(x +3) = 0$

$ <=> (\sqrt{x^{2} + 1} - \sqrt{x + 3})(\sqrt{x^{2} + 1} - 2\sqrt{x + 3}) = 0$

- TH1 $: \sqrt{x^{2} + 1} - \sqrt{x + 3}  = 0$

$ <=> x^{2} + 1 = x + 3 <=> x^{2} - x - 2 = 0$

$ <=> x = - 1; x = 2 (TM)$

- TH2 $: \sqrt{x^{2} + 1} - 2\sqrt{x + 3}  = 0$

$ <=> x^{2} + 1 = 4(x + 3) <=> x^{2} - 4x - 11 = 0$

$ <=> x = 2 - \sqrt{15}; x = 2 + \sqrt{15}(TM)$