[toán 12]mũ

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xuanquynh97

Ta có $y=e^{cosx}$

\Rightarrow $y^'=-sinxe^{cosx}; y^{''}=-cosxe^{cosx}+sin^2xe^{cosx} $

Do đó $y^'sinx+ycosx+y"= -sin^2xe^{cosx}+cosxe^{cosx} -cosxe^{cosx}+sin^2xe^{cosx}=0$

 
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