已知等差数列\(\left\{ {{a_n}} \right\}\)满足\({a_4} = {a_1} + {a_2}\)，且\({a_1} \ne 0\)，则\(\frac{{{a_5}}}{{{a_1} + {a_3}}} = \)___．

因为\({a_4} = {a_1} + {a_2}\)，所以\({a_1} + 3d = {a_1} + {a_1} + d\)，即\({a_1} = 2d\).

因为\({a_1} \ne 0\)，则\(d \ne 0\)，

所以\(\frac{{{a_5}}}{{{a_1} + {a_3}}} = \frac{{{a_1} + 4d}}{{{a_1} + {a_1} + 2d}} = \frac{{6d}}{{6d}} = 1\).

故答案为：1

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