If $x$ and $y$ are positive integers less than 10, which of the following COULD be true?
i) $\sqrt{x} + \sqrt{y} = \sqrt{x+y}$
ii) $y\sqrt{x} = x\sqrt{y}$
iii) $\sqrt{x} - \sqrt{y} = \sqrt{x-y}$

This one is tricky! I initially thought all three could be true, but then I remembered x and y are positive integers less than 10. That really limits the possibilities. I had to test each option.

For iii, I think it works when x > y, but if x < y then the right side is imaginary. So we need x > y. Also, after squaring, we get -2√(xy) = -2√(xy)? Wait, let me redo: (√x - √y)^2 = x + y - 2√(xy), and right side squared is x - y. So x + y - 2√(xy) = x - y => 2y = 2√(xy) => y = √(xy) => y^2 = xy => y = x (since y>0). So x=y. But then left side is 0, right side is 0. So iii works when x=y. But if x=y, then it's 0=0. So iii could be true. But wait, if x=y, then i becomes 2√x = √(2x), which is not true generally. So i is false. ii: y√x = x√y => square both sides: y^2 x = x^2 y => xy(y - x) = 0 => since x,y>0, y=x. So ii also works when x=y. So both ii and iii could be true when x=y. But wait, the question says 'could be true' meaning there exists some x,y? So if x=y=1, then ii: 1*1 = 1*1 works, iii: 1-1=0, √0=0 works. So both ii and iii could be true. i is never true? Let's check: √x+√y=√(x+y). Square: x+y+2√(xy)=x+y => 2√(xy)=0 => xy=0, impossible. So i is false. So answer should be D. But wait, the choices: D is ii and iii only. So that's correct. But is there any other possibility? For ii, we got x=y. For iii, we also got x=y. So only when x=y. So both ii and iii could be true simultaneously. So D is correct.

I got D. But I wasted time trying to find a case for i. It's impossible because squaring gives 2√(xy)=0, so x or y would have to be 0, but they're positive. So i is out. ii and iii both require x=y, which is allowed since they're positive integers less than 10. So D.