Math Dump /sym/6c3138

id=1 · host=fr.findroms.cloud · 2026-10-19 00:43Z

d/dy [y^8] = 8y^7

N₂ + 3H₂ ⇌ 2NH₃

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∑_{k=1}^{n} k = n(n+1)/2

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\lim_{x\to 0} \frac{\sin x}{x} = 1

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

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10γ² + 8γ + 7 = 0

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d/dn [n^3] = 3n^2

∇²γ = 0

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

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6t² + 4t + 0 = 0

e^{i\pi} + 1 = 0

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∑_{k=1}^{n} k = n(n+1)/2

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H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

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∂z/∂φ = 7z

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det| 7 4 ; 4 8 | = 40

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∫₀^∞ e^(-α²) dα = √π / 2

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