Đáp án:

 VY ngủ ngon.

Giải thích các bước giải:

 ĐKXĐ: `x ge 1/2`
`\sqrt{10x -5}+ \sqrt{5x^2 +5} = 3\sqrt{x^2+ 2x}`
`\sqrt{5(2x- 1)} +\sqrt{5(x^2 +1)} = 3\sqrt{x^2 +2x}`
`5(2x -1)+ 5(x^2 +1) + 2. 5\sqrt{(2x -1)(x^2+ 1)} = 9(x^2+ 2x)`
`5x^2 +10x +10\sqrt{(2x- 1)(x^2 +1)} = 9x^2+ 18x`
`4x^2+ 8x -10\sqrt{(2x -1)(x^2+ 1)} = 0`
`2x^2 +4x -5\sqrt{(2x- 1)(x^2 +1)} = 0`
`2x^2 +4x -5\sqrt{(2x- 1)(x^2 +1)} = 0`
`2(x^2 +1) -4\sqrt{x^2+ 1}\sqrt{2x -1} -\sqrt{x^2+ 1}\sqrt{2x -1}+ 2(2x -1) = 0`
`2\sqrt{x^2 +1}(\sqrt{x^2+ 1} -2\sqrt{2x -1}) -\sqrt{2x- 1}(\sqrt{x^2 +1}- 2\sqrt{2x -1}) = 0`
`(\sqrt{x^2+ 1} -2\sqrt{2x -1})(2\sqrt{x^2 +1}- \sqrt{2x -1}) = 0`
`+)\sqrt{x^2 +1}- 2\sqrt{2x -1} = 0`
`\sqrt{x^2+ 1} = 2\sqrt{2x- 1}`
`x^2 +1 = 4(2x -1)`
`x^2- 8x +5 = 0`
`(x^2 -8x+ 16) = 11`
`iff (x -4)^2 = (\sqrt{11})^2`
`@x- 4 = \sqrt{11} \iff x = 4 +\sqrt{11}` (t/m)
`@x -4= -\sqrt{11} \iff x = 4- \sqrt{11}` (t/m)
`+) 2\sqrt{x^2+ 1} -\sqrt{2x -1} = 0`
`2\sqrt{x^2 +1} = \sqrt{2x- 1}`
`4(x^2+ 1) = 2x -1`
`4x^2 -2x +5 = 0`
`(4x^2 -2x+ 1/4) +19/4 = 0`
Do `(4x^2 -2x+ 1/4) ge 0 ->(4x^2 -2x+ 1/4) +19/4 ge 19/4 >0 -> x in emptyset`
Vậy PT cs nghiệm : `x in { 4 -\sqrt{11}; 4 +\sqrt{11} }`