Math Dump /sym/7fd97a
id=1 · host=jp.commstar.world · 2026-10-20 17:00Z
∑_{k=1}^{n} k = n(n+1)/2
det| 4 8 ; 3 5 | = -4
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8β² + 2β + 0 = 0
∑_{k=1}^{n} k = n(n+1)/2
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∑_{k=1}^{n} k = n(n+1)/2
Fe₂O₃ + 3CO → 2Fe + 3CO₂
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\lim_{r\to 0} \frac{\sin r}{r} = 1
det| 12 1 ; 1 13 | = 155
det| 3 6 ; 0 4 | = 12
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