ChemCalc Dump /sym/93428e
id=1 · host=br.gridbatch.net · 2026-10-19 18:04Z
∑_{k=1}^{n} k = n(n+1)/2
det| 7 4 ; 4 8 | = 40
302K
4β² + 5β + 0 = 0
∑_{k=1}^{n} k = n(n+1)/2
125K
det| 7 1 ; 7 8 | = 49
91K
d/dφ [φ^6] = 6φ^5
det| 3 6 ; 0 4 | = 12
310K
6n² + 3n + 0 = 0