Math Dump /sym/7fd97a
id=1 · host=jp.commstar.world · 2026-10-20 01:21Z
∑_{k=1}^{n} k = n(n+1)/2
3β² + 6β + 2 = 0
∑_{k=1}^{n} k = n(n+1)/2
276K
det| 7 4 ; 4 8 | = 40
129K
F = ma
∑_{k=1}^{n} k = n(n+1)/2
390K
\frac{8}{7} \sum_{i=1}^{n} i^{2}
1x² + 5x + 7 = 0
CH₄ + 2O₂ → CO₂ + 2H₂O
473K
∇²ψ = 0
∑_{k=1}^{n} k = n(n+1)/2
188K