9. $1, \sqrt{7}, \sqrt{13}, \sqrt{19}, 5$ Let $[x]=n$, then $x^{2}-6 n+5=0, 6 n=x^{2}+5>0, n>0$. From $[x]=n \leqslant x$ we know, $n^{2}+5 \leqslant 6 n<(n+1)^{2}+5$, thus $n \in[1,5]$. Substituting $n=1,2,3,4,5$ respectively will do.
9. The real solutions of the equation $x^{2}-6[x]+5=0$ are

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