ChemCalc Dump /sym/93428e
id=1 · host=br.gridbatch.net · 2026-10-19 18:22Z
∑_{k=1}^{n} k = n(n+1)/2
det| 1 4 ; 3 2 | = -10
312K
4β² + 5β + 0 = 0
det| 7 4 ; 4 8 | = 40
125K
d/dφ [φ^6] = 6φ^5
det| 12 1 ; 5 13 | = 151
310K
det| 2 7 ; 7 3 | = -43
68K
∑_{k=1}^{n} k = n(n+1)/2
364K