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若x1,x2,x3,x4,x5为互不相等的正奇数,且满足(2011-x1)(2011-x2)(2011-x3)(2011-x4)(2011-x5)=24^2
那么,(x1)^2+(x2)^2+(x3)^2+(x4)^2+(x5)^2的末尾数字==?
人气:345 ℃ 时间:2020-10-01 18:43:55
解答
(2005-x1)(2005-x2)(2005-x3)(2005-x4)(2005-x5)=24²,而24²=2×(-2)×4×6×(-6),(2005-x1)²+(2005-x2)²+(2005-x3)²+(2005-x4)²+(2005-x5)²=2²+(-2)²...
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