로그아웃한 편집자를 위한 문서 더 알아보기
첫째날 Luminescent께서 말씀하시기를 F = A Π V ϵ ∫ ϕ ν ∫ β α A ( r ) σ 2 q 1 + q 2 sin ϕ tan ϕ n h + m v 2 − ν {\displaystyle F={\mathcal {A}}_{\mathit {\Pi }}\,V_{\epsilon }\int _{\phi }^{\nu }\int _{\beta }^{\alpha }A(r)\,\sigma ^{2}{\frac {q_{1}+q_{2}\sin \phi \tan \phi }{nh+m\,v^{2}-\nu }}} = 1 2 ζ ∮ ℵ K ∑ α = 0 ∞ H + I 1 I 2 e V + tan ϕ csc ϕ e + ∫ ∫ r ( A ) {\displaystyle ={\frac {1}{2}}\zeta \oint _{\aleph }^{K}\sum _{\alpha =0}^{\infty }{\frac {H+I_{1}\,I_{2}}{\sqrt[{e}]{eV+\tan \phi \csc \phi }}}+\int \int r(A)} α β | sin α cos β tan γ Ψ + q c o s e c τ sec π h sin ξ sin X e V sin θ | + ∮ ∫ Δ m Γ {\displaystyle {\frac {\alpha }{\beta }}\,{\begin{vmatrix}\sin \alpha &\cos \beta &\tan \gamma \\{\sqrt {{\mathit {\Psi }}+q}}&\mathrm {cosec} \,\tau &\sec \pi \\h\sin \xi &\sin X&eV\sin \theta \end{vmatrix}}+\oint \int {\sqrt[{\mathit {\Gamma }}]{{\mathit {\Delta }}m}}} = L m ( 91 235 U ) λ 0 μ 1 η + ∫ ∫ r ( A ) k V ( x , y ) h sin ( Φ ) Φ = 0 {\displaystyle =Lm(\,{\begin{smallmatrix}91\,\,\,\\235\end{smallmatrix}}U)\,\lambda _{0}\,\mu _{1}\eta +\int \int r(A){\frac {kV(x,\,y)}{h\sin({\mathit {\Phi }})}}_{{\mathit {\Phi }}=0}} ∫ ∫ ∫ 2 4 α μ e + F ( f 0 f 1 f 2 f 3 ) r ( A ) {\displaystyle \int \int \int {\frac {{\begin{smallmatrix}2\\4\end{smallmatrix}}\alpha }{{\sqrt[{e}]{\mu }}+F(f_{0}\,f_{1}\,f_{2}\,f_{3})}}\,r(A)} = L i m q → ∞ ∫ Δ m V Γ ∫ β α ϕ 2 R e ∫ Φ K e 2 m r W e 2 {\displaystyle ={\underset {q\to \infty }{\mathrm {Lim} }}\int {\sqrt[{\mathit {\Gamma }}]{{\mathit {\Delta }}mV}}\int _{\beta }^{\alpha }\phi ^{2}\,{\begin{matrix}R\\e\end{matrix}}\int {\mathit {\Phi }}{\sqrt {\frac {K\,e^{2}}{m_{r}}}}\,We^{2}} = h c λ = h ν = E {\displaystyle =h{\frac {c}{\lambda }}=h\nu =E} 라고 하시어 발광하셨느니라