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Question-212006




Question Number 212006 by RojaTaniya last updated on 26/Sep/24
Answered by a.lgnaoui last updated on 26/Sep/24
  posons  z=x+100    (((z−2)^5 +(z+2)^5 )/((z−1)^5 +(z+1)^5 ))=((2z^5 +80z^3 +160z)/(2z^5 +20z^3 +10z))  =((z^4 +40z^2 +80)/(z^4 +10z^2 +5))=((16^2 )/(11^2 −60))=((256)/(61))  alors  61z^4 +2440z^2 +4880=256z^4 +2560z^2 +128  256z^4 2560z^2 +1280         13z^4 +8z^2 −240=0         16+3120=56^2          z^2 =4     z={−2,+2}          { ((x=−98)),((x=−102)) :}
$$\:\:\mathrm{posons}\:\:\boldsymbol{\mathrm{z}}=\boldsymbol{\mathrm{x}}+\mathrm{100} \\ $$$$\:\:\frac{\left(\boldsymbol{\mathrm{z}}−\mathrm{2}\right)^{\mathrm{5}} +\left(\boldsymbol{\mathrm{z}}+\mathrm{2}\right)^{\mathrm{5}} }{\left(\boldsymbol{\mathrm{z}}−\mathrm{1}\right)^{\mathrm{5}} +\left(\boldsymbol{\mathrm{z}}+\mathrm{1}\right)^{\mathrm{5}} }=\frac{\mathrm{2}\boldsymbol{\mathrm{z}}^{\mathrm{5}} +\mathrm{80}\boldsymbol{\mathrm{z}}^{\mathrm{3}} +\mathrm{160}\boldsymbol{\mathrm{z}}}{\mathrm{2}\boldsymbol{\mathrm{z}}^{\mathrm{5}} +\mathrm{20}\boldsymbol{\mathrm{z}}^{\mathrm{3}} +\mathrm{10}\boldsymbol{\mathrm{z}}} \\ $$$$=\frac{\boldsymbol{\mathrm{z}}^{\mathrm{4}} +\mathrm{40}\boldsymbol{\mathrm{z}}^{\mathrm{2}} +\mathrm{80}}{\boldsymbol{\mathrm{z}}^{\mathrm{4}} +\mathrm{10}\boldsymbol{\mathrm{z}}^{\mathrm{2}} +\mathrm{5}}=\frac{\mathrm{16}^{\mathrm{2}} }{\mathrm{11}^{\mathrm{2}} −\mathrm{60}}=\frac{\mathrm{256}}{\mathrm{61}} \\ $$$$\boldsymbol{\mathrm{alors}} \\ $$$$\mathrm{61}\boldsymbol{\mathrm{z}}^{\mathrm{4}} +\mathrm{2440}\boldsymbol{\mathrm{z}}^{\mathrm{2}} +\mathrm{4880}=\mathrm{256}\boldsymbol{\mathrm{z}}^{\mathrm{4}} +\mathrm{2560}\boldsymbol{\mathrm{z}}^{\mathrm{2}} +\mathrm{128} \\ $$$$\mathrm{256}\boldsymbol{\mathrm{z}}^{\mathrm{4}} \mathrm{2560}\boldsymbol{\mathrm{z}}^{\mathrm{2}} +\mathrm{1280} \\ $$$$ \\ $$$$\:\:\:\:\:\mathrm{13z}^{\mathrm{4}} +\mathrm{8z}^{\mathrm{2}} −\mathrm{240}=\mathrm{0} \\ $$$$ \\ $$$$\:\:\:\:\:\mathrm{16}+\mathrm{3120}=\mathrm{56}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\boldsymbol{\mathrm{z}}^{\mathrm{2}} =\mathrm{4}\:\:\:\:\:\boldsymbol{\mathrm{z}}=\left\{−\mathrm{2},+\mathrm{2}\right\} \\ $$$$\:\: \\ $$$$\:\:\:\begin{cases}{\mathrm{x}=−\mathrm{98}}\\{\mathrm{x}=−\mathrm{102}}\end{cases} \\ $$$$ \\ $$$$ \\ $$

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