if $x = \sqrt{x+1} + \sqrt{x+2} + \sqrt{x+3}$ then x =? I have got a new question from my friend and it made me nervous:
$$\text{ If }\,x = \sqrt{x+1} + \sqrt{x+2} + \sqrt{x+3}, \text{ then }\,x = \;?$$
A lot of thank you to all comments.

$$x=\sqrt{x+1}+\sqrt{x+2}+\sqrt{x+3}$$
$$x-\sqrt{x+1}=\sqrt{x+2}+\sqrt{x+3}$$
$$x^2+x+1-2x\sqrt{x+1}=2x+5+2\sqrt{x^2+5x+6}$$
$$x^2-x-4=2(\sqrt{x^3+x^2}+\sqrt{x^2+5x+6})$$
$$x^4-2x^3-7x^2+8x+16=4(x^3+2x^2+5x+6+2\sqrt{x^5+6x^4+11x^3+6x^2})$$
$$x^4-6x^3-15x^2-12x-8=8\sqrt{x^5+6x^4+11x^3+6x^2}$$
$$x^8-12x^7+6x^6+156x^5+353x^4+456x^3+384x^2+192x+64=64x^5+384x^4+704x^3+384x^2$$
$$x^8-12x^7+6x^6+92x^5-31x^4-248x^3+192x+64=0$$
This equation has $6$ real roots, but only the one I mentioned in my comment fulfills the original equation. Therefore, it is the only real solution.
