Math Dump /sym/6c3138

id=1 · host=fr.findroms.cloud · 2026-10-18 15:17Z

11n² + 5n + 4 = 0

e^{i\pi} + 1 = 0

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N₂ + 3H₂ ⇌ 2NH₃

438K

∫₀^∞ e^(-x²) dx = √π / 2

\binom{n}{k} = \frac{n!}{k!(n-k)!}

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det| 2 6 ; 0 3 | = 6

411K

4φ² + 6φ + 2 = 0

iħ ∂ψ/∂t = Ĥψ

448K

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

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∫₀^∞ e^(-φ²) dφ = √π / 2

∑_{k=1}^{n} k = n(n+1)/2

459K

8ψ² + 7ψ + 0 = 0

406K

det| 5 2 ; 3 6 | = 24

181K

∑_{k=1}^{n} k = n(n+1)/2

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