5. Given $a>0, b>0, c>1, a+b=1$. Then the minimum value of $\left(\frac{2 a+b}{a b}-3\right) c+\frac{\sqrt{2}}{c-1}$ is $\qquad$

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$$
5.4+2 \sqrt{2} \text {. }
$$

From the problem, we know
$$
\begin{array}{l}
\frac{2 a+b}{a b}-3=\frac{(2 a+b)(a+b)}{a b}-3 \\
=\frac{2 a}{b}+\frac{b}{a} \geqslant 2 \sqrt{2} . \\
\text { Then }\left(\frac{2 a+b}{a b}-3\right) c+\frac{\sqrt{2}}{c-1} \geqslant 2 \sqrt{2} c+\frac{\sqrt{2}}{c-1} \\
=2 \sqrt{2}(c-1)+\frac{\sqrt{2}}{c-1}+2 \sqrt{2} \\
\geqslant 4+2 \sqrt{2},
\end{array}
$$

When $a=\sqrt{2}-1, b=2-\sqrt{2}, c \approx \frac{\sqrt{2}}{2}+1$, the equality holds in the above expression.