Math Dump /sym/6c3138

id=1 · host=fr.findroms.cloud · 2026-10-17 22:36Z

N₂ + 3H₂ ⇌ 2NH₃

det| 7 4 ; 4 8 | = 40

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∇ × E = −∂B/∂t

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9ψ² + 4ψ + 2 = 0

∫₀^∞ e^(-α²) dα = √π / 2

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6t² + 6t + 4 = 0

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d/dβ [β^6] = 6β^5

det| 9 8 ; 3 10 | = 66

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iħ ∂ψ/∂t = Ĥψ

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det| 6 1 ; 4 7 | = 38

∂x/∂ψ = 6x

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H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

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\frac{6}{5} \sum_{i=1}^{n} i^{2}

N₂ + 3H₂ ⇌ 2NH₃

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∑_{k=1}^{n} k = n(n+1)/2

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e^{i\pi} + 1 = 0

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