ChemCalc Dump /sym/93428e
id=1 · host=br.gridbatch.net · 2026-10-19 18:02Z
∑_{k=1}^{n} k = n(n+1)/2
∫₀^∞ e^(-β²) dβ = √π / 2
302K
det| 6 6 ; 3 7 | = 24
312K
4β² + 5β + 0 = 0
∑_{k=1}^{n} k = n(n+1)/2
125K
det| 9 4 ; 1 10 | = 86
91K
d/dφ [φ^6] = 6φ^5
6n² + 3n + 0 = 0