Symbolic Index /sym/ea4bfa
id=1 · host=au.commstar.world · 2026-10-20 14:19Z
∑_{k=1}^{n} k = n(n+1)/2
∫₀^∞ e^(-t²) dt = √π / 2
52K
det| 12 5 ; 4 13 | = 136
det| 11 5 ; 0 12 | = 132
90K
det| 2 7 ; 7 3 | = -43
91K
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
∑_{k=1}^{n} k = n(n+1)/2
100K
det| 4 9 ; 7 5 | = -43
det| 9 4 ; 1 10 | = 86
137K
det| 1 1 ; 3 2 | = -1
∑_{k=1}^{n} k = n(n+1)/2
262K
∫₀^∞ e^(-φ²) dφ = √π / 2
196K
det| 3 3 ; 4 4 | = 0
218K