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x-V-J-x-ustx-zac-2-x-x-ustx-0-x-1-v-pi-2-L-l-x-0-v-X-2x-2-1-




Question Number 222523 by hu last updated on 29/Jun/25
x+V−J(x)((ustx)/(zac^2 x))=x−((ustx_0 )/( (√x)ψ+ζ(((−1+v%)/(π+2))+L_(l(x→0)) ^(  v%) X_(2x^2 ) ^( 1) )))
$${x}+{V}−{J}\left({x}\right)\frac{{ustx}}{{zac}^{\mathrm{2}} {x}}={x}−\frac{{ustx}_{\mathrm{0}} }{\:\sqrt{{x}}\psi+\zeta\left(\frac{−\mathrm{1}+{v\%}}{\pi+\mathrm{2}}+\mathscr{L}_{{l}\left({x}\rightarrow\mathrm{0}\right)} ^{\:\:{v\%}} {X}_{\mathrm{2}{x}^{\mathrm{2}} } ^{\:\mathrm{1}} \right)} \\ $$
Answered by wewji12 last updated on 29/Jun/25
≪^(SL{Fv_(QwR) }) ∞^(vcos(θ)) −μζJ^V (uTt_0 )^(12345)
$$\ll^{{S}\mathcal{L}\left\{{Fv}_{{QwR}} \right\}} \infty^{{v}\mathrm{cos}\left(\theta\right)} −\mu\zeta{J}^{{V}} \left({uTt}_{\mathrm{0}} \right)^{\mathrm{12345}} \\ $$
Answered by MathematicalUser2357 last updated on 09/Jul/25
this is me
$$\mathrm{this}\:\mathrm{is}\:\mathrm{me} \\ $$
Commented by MathematicalUser2357 last updated on 09/Jul/25

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