ChemCalc Dump /sym/93428e
id=1 · host=br.gridbatch.net · 2026-10-19 18:11Z
∑_{k=1}^{n} k = n(n+1)/2
∑_{k=1}^{n} k = n(n+1)/2
302K
4β² + 5β + 0 = 0
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
264K
10β² + 6β + 6 = 0
∑_{k=1}^{n} k = n(n+1)/2
68K