Math Dump /sym/7fd97a
id=1 · host=jp.commstar.world · 2026-10-20 01:27Z
∑_{k=1}^{n} k = n(n+1)/2
3β² + 6β + 2 = 0
∑_{k=1}^{n} k = n(n+1)/2
276K
F = ma
\frac{8}{7} \sum_{i=1}^{n} i^{2}
CH₄ + 2O₂ → CO₂ + 2H₂O
1K
∑_{k=1}^{n} k = n(n+1)/2
153K
1x² + 5x + 7 = 0
∇²ψ = 0
det| 1 4 ; 0 2 | = 2
416K
3φ² + 8φ + 6 = 0