ChemCalc Dump /sym/93428e
id=1 · host=br.gridbatch.net · 2026-10-19 18:04Z
∑_{k=1}^{n} k = n(n+1)/2
∑_{k=1}^{n} k = n(n+1)/2
312K
4β² + 5β + 0 = 0
det| 9 1 ; 7 10 | = 83
125K
det| 5 2 ; 3 6 | = 24
91K
d/dφ [φ^6] = 6φ^5
det| 7 4 ; 4 8 | = 40
310K
6n² + 3n + 0 = 0
det| 5 7 ; 2 6 | = 16
39K