Math Dump /sym/7fd97a
id=1 · host=jp.commstar.world · 2026-10-20 17:03Z
∑_{k=1}^{n} k = n(n+1)/2
det| 3 7 ; 2 4 | = -2
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8β² + 2β + 0 = 0
∑_{k=1}^{n} k = n(n+1)/2
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C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
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∑_{k=1}^{n} k = n(n+1)/2
det| 7 4 ; 4 8 | = 40
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C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
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