在等比数列 \(\left\{ {{a_n}} \right\}\) 中，前 \(10\) 项和是 \(10\) ，\({a_1} - {a_2} + {a_3} - {a_4} + {a_5} - {a_6} + {a_7} - {a_8} + {a_9} - {a_{10}} = 5\) ，则数列 \(\left\{ {{a_n}} \right\}\) 的公比 \(q = \) ___.

由题意可得 \(\left\{ {\begin{array}{*{20}{l}}
{{a_1} - {a_2} + {a_3} - {a_4} + {a_5} - {a_6} + {a_7} - {a_8} + {a_9} - {a_{10}} = 5} \\\
{{a_1} + {a_2} + {a_3} + {a_4} + {a_5} + {a_6} + {a_7} + {a_8} + {a_9} + {a_{10}} = 10}
\end{array}} \right.\) ，

得 \(\left\{ {\begin{array}{*{20}{l}}
{{a_1} + {a_3} + {a_5} + {a_7} + {a_9} = \frac{{15}}{2}} \\\
{{a_2} + {a_4} + {a_6} + {a_8} + {a_{10}} = \frac{5}{2}}
\end{array}} \right.\) ，\(\because q = \frac{{{a_2}}}{{{a_1}}} = \frac{{{a_4}}}{{{a_3}}} = \frac{{{a_6}}}{{{a_5}}} = \frac{{{a_8}}}{{{a_7}}} = \frac{{{a_{10}}}}{{{a_9}}}\) ，

由连比定理得 \(q = \frac{{{a_2} + {a_4} + {a_6} + {a_8} + {a_{10}}}}{{{a_1} + {a_3} + {a_5} + {a_7} + {a_9}}} = \frac{5}{2} \div \frac{{15}}{2} = \frac{5}{2} \times \frac{2}{{15}} = \frac{1}{3}\) ，故答案为 \(\frac{1}{3}\) .

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