jmc

We guess that $\sqrt{2}-\sqrt{5}$ is also a root of $P(x).$ In that case, $P(x)$ must be divisible by the polynomial $$(x-(\sqrt{2}+\sqrt{5}))(x-(\sqrt{2}-\sqrt{5}))=x^2-2x\sqrt{2}-3.$$Now we see that if we multiply this polynomial by $x^2+2x\sqrt{2}-3,$ we get a polynomial with rational coefficients: $$(x^2-2x\sqrt{2}-3)(x^2+2x\sqrt{2}-3)=x^4-14x^2+9.$$Therefore, $P(x)=x^4-14x^2+9,$ and so $P(1)=1-14+9=\boxed{-4}.$