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
det| 1 2 ; 7 2 | = -12
125K
d/dφ [φ^6] = 6φ^5
∫₀^∞ e^(-ω²) dω = √π / 2
17K
6n² + 3n + 0 = 0
det| 11 9 ; 5 12 | = 87
264K
10β² + 6β + 6 = 0
4β² + 3β + 0 = 0
F = ma
d/dα [α^1] = 1α^0