All The Digits

First, let's get some facts straight:

1) The fact the $\sqrt{A},\sqrt{B},\sqrt{C}$ are rationals means that $A,B,C$ are perfect squares. Hence they can only be $1,4,9$ .

2)The fact that $\dfrac{H}{I}$ is a terminating decimal means that $I$ is either 2,5 or 8.

Now assume WLOG that $A=1,B=4,C=9$ .Then we are left with the following digits: $\{2,3,5,6,7,8\}$ Now since $\dfrac{D}{E}$ and $\dfrac{F}{G}$ are integers, then $D$ and $F$ are integral multiples of $E$ and $G$ respectively. Observe that out of our remaining digits, only 2 and 3 have single non-digits multiples, namely 6 and 8.Note that $D,F\neq 2,3,5,7$ since these donot have any single digits factors apart from themselves. Therefore, $D$ and $F$ can only be 6 or 8.WLOG assume that $D=6$ .Then $F=8$ and therefore $G=2$ and $E=3$ .Now we are left with the following digits: $\{5,7\}$ Now, since $\dfrac{H}{I}$ is a terminating decimal necessarily implies that $H=7$ and $I=5$ . Therefore $\boxed{H=7}$