Đkxđ: `x ge 0`
`x^2 +5x+ 1 = 4\sqrt{x}(x +1)`
`(x^2+ 2x +1) -4(x +1)\sqrt{x}+ 3x = 0`
`(x +1)^2- (x +1)\sqrt{x} -3(x+ 1)\sqrt{x}+ 3x= 0`
`(x+ 1)(x +1 -\sqrt{x}) - 3\sqrt{x}(x +1- \sqrt{x}) =0`
`(x -\sqrt{x}+ 1)(x- 3\sqrt{x} +1) = 0`
`+) x -\sqrt{x} +1 = 0`
`-> x -2 .\sqrt{x} .(1/2)+ 1/4 +3/4 =0`
`-> (\sqrt{x} -1/2)^2+ 3/4 =0`
Vì `(\sqrt{x} -1/2)^2 ge 0 -> (\sqrt{x} -1/2)^2+ 3/4 ge 3/4 >0 ->` Vô lí
`-> x in \emptyset`
`+) x- 3\sqrt{x} +1 =0`
`x -2. \sqrt{x} .(3/2) +9/4 = 5/4`
`(\sqrt{x}- 3/2)^2 = (\sqrt{5}/2)^2`
`@ \sqrt{x} -3/2= \sqrt{5}/2 `
`\iff \sqrt{x} = (3 +\sqrt{5})/2`
`\iff x= ((3+ \sqrt{5})/2)^2 = (14 +6\sqrt{5})/4= (7 +3\sqrt{5})/2` (t/m)
`@ \sqrt{x}- 3/2 = -\sqrt{5}/2`
`\iff \sqrt{x}= (3 -\sqrt{5})/2`
`\iff x = ((3 -\sqrt{5})/2)^2= (14 -6\sqrt{5})/4 = (7- 3\sqrt{5})/2` (t/m)
Vậy PT có nghiệm:  `x in {(7 -3\sqrt{5})/2; (7+ 3\sqrt{5})/2 }`