ChemCalc Dump /sym/93428e
id=1 · host=br.gridbatch.net · 2026-10-19 18:06Z
∑_{k=1}^{n} k = n(n+1)/2
4β² + 5β + 0 = 0
det| 10 9 ; 0 11 | = 110
125K
d/dφ [φ^6] = 6φ^5
∑_{k=1}^{n} k = n(n+1)/2
17K
6n² + 3n + 0 = 0
Fe₂O₃ + 3CO → 2Fe + 3CO₂
311K
det| 9 7 ; 3 10 | = 69
264K
∑_{k=1}^{n} k = n(n+1)/2
364K