5. Given that $a, b$ are positive numbers, and $\frac{a+b}{a^{2}+a b+b^{2}}=\frac{4}{49}$, then the value of $a+b$ is
5. 16 Detailed Explanation: Let $a+b=4 k\left(k \in \mathbf{N}^{*}\right), a^{2}+a b+b^{2}=49 k$,
$a b=16 k^{2}-49 k$, thus $a, b$ are the two positive roots of the equation $x^{2}-4 k x+\left(16 k^{2}-49 k\right)=0$, by $\Delta \geqslant 0 \Rightarrow 0 \leqslant k \leqslant \frac{49}{12} \Rightarrow$ $k=1,2,3,4$ substituting and checking yields $k=4$.