Math Dump /sym/7fd97a
id=1 · host=jp.commstar.world · 2026-10-20 01:16Z
∑_{k=1}^{n} k = n(n+1)/2
∑_{k=1}^{n} k = n(n+1)/2
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3β² + 6β + 2 = 0
F = ma
det| 11 4 ; 5 12 | = 112
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∑_{k=1}^{n} k = n(n+1)/2
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∑_{k=1}^{n} k = n(n+1)/2
338K
det| 7 4 ; 4 8 | = 40
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