The only foursomes that multiply to one less than a square are: $$1\times2\times3\times4=5^2-1\\ 1\times2\times3\times8=7^2-1\\ 1\times2\times4\times6=7^2-1\\ 1\times4\times8\times9=17^2-1\\ 2\times3\times6\times8=17^2-1$$
Clearly $x\ne5,7$: else we can't find four of these numbers adding to $x-1$. So $x=17$.
Suppose the four numbers used in the bottom oval are $1, 4, 8, 9$. That leaves $2, 3, 6$. For the upper-left oval to have $2$ less than than the upper-right, we need the upper-left to have $2, 8$ or $4, 6$ to the other's $3, 9$, or to have $3, 9$ to the other's $6, 8$. None of those possibilities has numbers available for the middle-top two segments that will allow the sums to be $16$ and $18$.
So the four numbers used in the bottom oval are $2, 3, 6, 8$. That leaves $1, 4, 9$. For the upper-left oval to have $2$ less than than the upper-right, we need the upper-left to have $1, 8$ to the other's $2, 9$, or to have $3, 4$ to the other's $1, 8$, or to have $4, 6$ to the other's $3, 9$. Only the first of those possibilities has numbers available for the middle-top two segments that will allow the sums to be $16$ and $18$: $3$ and $4$. So we obtain