ChemCalc Dump /sym/93428e
id=1 · host=br.gridbatch.net · 2026-10-19 18:05Z
∑_{k=1}^{n} k = n(n+1)/2
det| 5 6 ; 4 6 | = 6
302K
4β² + 5β + 0 = 0
det| 11 9 ; 0 12 | = 132
91K
d/dφ [φ^6] = 6φ^5
6n² + 3n + 0 = 0
det| 7 4 ; 4 8 | = 40
311K
∑_{k=1}^{n} k = n(n+1)/2
68K