3. For the equation with respect to $x$
$$
\sqrt{2}|x-k|=k \sqrt{x}
$$

to have two distinct real roots in the interval $[k-1, k+1]$, the range of the real number $k$ is
3. $00$.
$$
\text { Let } f(x)=|x-k|, g(x)=\frac{\sqrt{2}}{2} k \sqrt{x} \text {. }
$$

When $01$, $\left\{\begin{array}{l}f(k-1) \geqslant g(k-1), \\ f(k+1) \geqslant g(k+1) .\end{array}\right.$
But $f(k+1)<g(k+1)$, no solution.