ChemCalc Dump /sym/93428e
id=1 · host=br.gridbatch.net · 2026-10-19 18:11Z
∑_{k=1}^{n} k = n(n+1)/2
∫₀^∞ e^(-z²) dz = √π / 2
312K
4β² + 5β + 0 = 0
d/dφ [φ^6] = 6φ^5
det| 10 4 ; 1 11 | = 106
17K
det| 12 1 ; 5 13 | = 151
310K
6n² + 3n + 0 = 0
det| 9 4 ; 1 10 | = 86
264K