ChemCalc Dump /sym/93428e
id=1 · host=br.gridbatch.net · 2026-10-19 18:59Z
∑_{k=1}^{n} k = n(n+1)/2
4β² + 5β + 0 = 0
d/dφ [φ^6] = 6φ^5
det| 5 9 ; 7 6 | = -33
17K
6n² + 3n + 0 = 0
∑_{k=1}^{n} k = n(n+1)/2
311K
det| 7 1 ; 7 8 | = 49
264K
det| 7 4 ; 4 8 | = 40
68K