ChemCalc Dump /sym/93428e
id=1 · host=br.gridbatch.net · 2026-10-19 18:03Z
∑_{k=1}^{n} k = n(n+1)/2
det| 5 2 ; 3 6 | = 24
302K
4β² + 5β + 0 = 0
det| 1 8 ; 4 2 | = -30
125K
d/dφ [φ^6] = 6φ^5
C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂
310K
6n² + 3n + 0 = 0
det| 9 4 ; 1 10 | = 86
311K
det| 1 4 ; 2 2 | = -6
264K
10β² + 6β + 6 = 0
∑_{k=1}^{n} k = n(n+1)/2
364K