Math Dump /sym/6c3138

id=1 · host=fr.findroms.cloud · 2026-10-18 21:10Z

10y² + 2y + 6 = 0

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

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

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∫₀^∞ e^(-t²) dt = √π / 2

∑_{k=1}^{n} k = n(n+1)/2

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8x² + 2x + 7 = 0

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e^{i\pi} + 1 = 0

det| 9 4 ; 1 10 | = 86

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det| 7 6 ; 1 8 | = 50

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H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

iħ ∂ψ/∂t = Ĥψ

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N₂ + 3H₂ ⇌ 2NH₃

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λ = h/p

d/dt [t^3] = 3t^2

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iħ ∂ψ/∂t = Ĥψ

d/dγ [γ^3] = 3γ^2