Math Dump /sym/7fd97a
id=1 · host=jp.commstar.world · 2026-10-20 01:17Z
∑_{k=1}^{n} k = n(n+1)/2
3β² + 6β + 2 = 0
det| 1 4 ; 3 2 | = -10
276K
det| 12 1 ; 4 13 | = 152
129K
F = ma
det| 9 4 ; 1 10 | = 86
390K
\frac{8}{7} \sum_{i=1}^{n} i^{2}
1x² + 5x + 7 = 0
∑_{k=1}^{n} k = n(n+1)/2
473K
∑_{k=1}^{n} k = n(n+1)/2
338K
∇²ψ = 0