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

`x\sqrt{x+1}+2·\sqrt{2x+1}·\sqrt{x-2}=\sqrt{2x+1}·\sqrt{x+1}+2x\sqrt{x-2}`

`x\sqrt{x+1}-\sqrt{2x+1}·\sqrt{x+1}+2·\sqrt{2x+1}·\sqrt{x-2}-2x\sqrt{x-2}=0`

`\sqrt{x+1}(x-\sqrt{2x+1})+2\sqrt{x-2}(\sqrt{2x+1}-x)=0`

`\sqrt{x+1}(x-\sqrt{2x+1})-2\sqrt{x-2}(x-\sqrt{2x+1})=0`

`(\sqrt{x+1}-2\sqrt{x-2})(x-\sqrt{2x+1})=0`

`Th1:`

`\sqrt{x+1}-2\sqrt{x-2}=0`

`\sqrt{x+1}=2\sqrt{x-2}`

`x+1=4(x-2)`

`x+1=4x-8`

`x-4x=-8-1`

`-3x=-9`

`x=3`

`Th2:`

`x-\sqrt{2x+1}=0`

`x=\sqrt{2x+1}`

`x^2=2x+1`

`x^2-2x-1=0`

`Δ'=(-1)^2-1·(-1)=2>0`

`⇒` pt có `2` nghiệm pb

`x_1=(-(-1)+\sqrt{2})/1=1+\sqrt{2}`

`x_1=(-(-1)-\sqrt{2})/1=1-\sqrt{2}`

Thử lại thấy `1-\sqrt{2}` không thỏa mãn

Vậy `x in {3;1+\sqrt{2}}`