Điều kiện: x ≥ 0.

\(\sqrt {x + 3} + \sqrt {3x + 1} = 2\sqrt x + \sqrt {2x + 2} \)

⇔ \(\sqrt {x + 3} - 2 + \sqrt {3x + 1} - 2 = 2\sqrt x - 2 + \sqrt {2x + 2} - 2\)

⇔ \[\frac{{x + 3 - 4}}{{\sqrt {x + 3} + 2}} + \frac{{3x + 1 - 4}}{{\sqrt {3x + 1} + 2}} = \frac{{4x - 4}}{{2\sqrt x - 2}} + \frac{{2x + 2 - 4}}{{\sqrt {2x + 2} - 2}}\]

⇔ \[\frac{{x - 1}}{{\sqrt {x + 3} + 2}} + \frac{{3\left( {x - 1} \right)}}{{\sqrt {3x + 1} + 2}} = \frac{{4\left( {x - 1} \right)}}{{2\sqrt x - 2}} + \frac{{2\left( {x - 1} \right)}}{{\sqrt {2x + 2} - 2}}\]

⇔ \[\left( {x - 1} \right)\left( {\frac{1}{{\sqrt {x + 3} + 2}} + \frac{3}{{\sqrt {3x + 1} + 2}} - \frac{4}{{2\sqrt x - 2}} - \frac{2}{{\sqrt {2x + 2} - 2}}} \right) = 0\]

Ta thấy: \[\frac{1}{{\sqrt {x + 3} + 2}} + \frac{3}{{\sqrt {3x + 1} + 2}} - \frac{4}{{2\sqrt x - 2}} - \frac{2}{{\sqrt {2x + 2} - 2}}\] > 0

Suy ra: x – 1 = 0 hay x = 1 (t/m).

Vậy x = 1.