1523| 7
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(x+1)(y+1)(z+1)=20,(x+2)(y+2)(z+2)=21,(x+3)(y+3)(z+3)=2021,求 (x+10)(y+10)(z+10) |
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发表于 2021-1-23 19:31
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发表于 2021-1-23 20:23
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-6 + 3 a - 3 b + c + 1/2 (22 - 5 a + 8 b - 3 c) n + 1/2 (-12 + a - 2 b + c) n^2 + n^3
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发表于 2021-1-24 12:59
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发表于 2021-1-24 18:18
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