I assume the first 1990 has to be 1989. Write $x-1991=y$ then $\sqrt{{y+1984}\over {1989}}+\sqrt{{y+1985}\over {1990}}+\sqrt{{y+1986}\over {1991}}=\sqrt{{y+2}\over 7}+\sqrt{{y+1}\over 6}+\sqrt{y\over 5}$ We have $y\ge 0$ for any solution. Note that ${{y+1984}\over {1989}}={{y+2}\over 7}\iff {{y-5}\over {1989}}={{y-5}\over 7}\iff y=5$ and the same for the other two combinations. For $y<5$ we have ${{y+1984}\over {1989}}>{{y+2}\over 7}$ and the same for the other two. Hence no equality. Same when we flip the inequalities. Hence equality iff $y=5$ and then $x=1996$.

alexheinis wrote: I assume the first 1990 has to be 1989.. Indeed, tnx for noticing