Mathématiques

Question


1+  1+1sur 1-1/5 le tout sur 1-1 sur 1+1/5 égal

1 Réponse

  • Bonjour,

    [tex]1+\dfrac{1+\dfrac{1}{1-\dfrac{1}{5}}}{1-\dfrac{1}{1+\dfrac{1}{5}}}=1+\dfrac{1+\dfrac{1}{\dfrac{5}{5}-\dfrac{1}{5}}}{1-\dfrac{1}{\dfrac{5}{5}+\dfrac{1}{5}}}\\\\\\1+\dfrac{1+\dfrac{1}{1-\dfrac{1}{5}}}{1-\dfrac{1}{1+\dfrac{1}{5}}}=1+\dfrac{1+\dfrac{1}{\dfrac{4}{5}}}{1-\dfrac{1}{\dfrac{6}{5}}}\\\\\\[/tex]

    [tex]1+\dfrac{1+\dfrac{1}{1-\dfrac{1}{5}}}{1-\dfrac{1}{1+\dfrac{1}{5}}}=1+\dfrac{1+1\times\dfrac{5}{4}}{1-1\times{\dfrac{5}{6}}}\\\\\\1+\dfrac{1+\dfrac{1}{1-\dfrac{1}{5}}}{1-\dfrac{1}{1+\dfrac{1}{5}}}=1+\dfrac{1+\dfrac{5}{4}}{1-{\dfrac{5}{6}}}[/tex]


    [tex]1+\dfrac{1+\dfrac{1}{1-\dfrac{1}{5}}}{1-\dfrac{1}{1+\dfrac{1}{5}}}=1+\dfrac{\dfrac{4}{4}+\dfrac{5}{4}}{\dfrac{6}{6}-{\dfrac{5}{6}}}\\\\\\1+\dfrac{1+\dfrac{1}{1-\dfrac{1}{5}}}{1-\dfrac{1}{1+\dfrac{1}{5}}}=1+\dfrac{\dfrac{9}{4}}{\dfrac{1}{6}}}[/tex]

    [tex]\\\\\\1+\dfrac{1+\dfrac{1}{1-\dfrac{1}{5}}}{1-\dfrac{1}{1+\dfrac{1}{5}}}=1+\dfrac{9}{4}\times6}\\\\\\1+\dfrac{1+\dfrac{1}{1-\dfrac{1}{5}}}{1-\dfrac{1}{1+\dfrac{1}{5}}}=1+\dfrac{54}{4}}[/tex]


    [tex]1+\dfrac{1+\dfrac{1}{1-\dfrac{1}{5}}}{1-\dfrac{1}{1+\dfrac{1}{5}}}=1+\dfrac{27}{2}\\\\1+\dfrac{1+\dfrac{1}{1-\dfrac{1}{5}}}{1-\dfrac{1}{1+\dfrac{1}{5}}}=\dfrac{2}{2}+\dfrac{27}{2}\\\\1+\dfrac{1+\dfrac{1}{1-\dfrac{1}{5}}}{1-\dfrac{1}{1+\dfrac{1}{5}}}=\dfrac{29}{2}[/tex]

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