ChemCalc Dump /sym/93428e
id=1 · host=br.gridbatch.net · 2026-10-19 18:57Z
∑_{k=1}^{n} k = n(n+1)/2
det| 3 7 ; 2 4 | = -2
312K
4β² + 5β + 0 = 0
∑_{k=1}^{n} k = n(n+1)/2
91K
d/dφ [φ^6] = 6φ^5
det| 5 6 ; 1 6 | = 24
310K
6n² + 3n + 0 = 0
det| 9 4 ; 1 10 | = 86
311K
det| 5 7 ; 2 6 | = 16
364K