Răspuns:

Formula pe care am folosit-o este :

[tex]a^{\frac{m}{n} } =\sqrt[n]{a^{m} }[/tex]

Rezolvati in multimea numerelor reale ecuatia:

[tex]\sqrt[n]{n} ^{\sqrt[-n]{n} } = n[/tex]

[tex]\sqrt[- n]{n} = n^-{\frac{1}{n} }[/tex]

[tex]\sqrt[n]{n} = n^{\frac{1}{n} }[/tex]

Deci vom avea asa:

[tex]\sqrt[n]{n}^{-\sqrt[n]{n} } = n^{\frac{1^ }^{n} }^{-\sqrt[n]{n} } }=[/tex][tex]n^{\frac{1}{n}^{n^{-\frac{1}{n} } =[/tex][tex]n^{\frac{1}{n^{2} } ^{-\frac{1}{n} } =[/tex][tex]n^{-\sqrt[n]{\frac{1}{n^{2} } } }[/tex](nu ne prea ajuta)

sau

[tex]\sqrt[n]{n}^{-\sqrt[n]{n} } = n[/tex]

[tex]\sqrt[n]{n} *ln \sqrt[n]{n} =[/tex] ㏑ n

[tex]\sqrt[n]{n}*({\frac{1}{n})[/tex] * ㏑ n = ㏑ n

[tex]\sqrt[n]{n} = n^{\frac{1}{n} } = n^{1}[/tex]⇒ [tex]\frac{1}{n} = 1[/tex]⇒ n ∈{-1 ,1}

Verificam:

Daca n = 1 atunci  [tex]\sqrt[n]{n}^{-\sqrt[n]{n} } = \sqrt[1]{1}^{-\sqrt[1]{1} }= 1^{-1} = \frac{1}{1^{1} } = 1[/tex]

Daca n= -1 atunci [tex]\sqrt[n]{n}^{-\sqrt[n]{n} } = \sqrt[-1]{-1}^{-\sqrt[-1]{-1} }=- 1^{-1} = \frac{1}{-1^{1} } = -1[/tex]

Deci solutiile ecuatie sunt:

n∈ {-1, 1}

Iar -1, 1 ∈ R

Sper ca te-am ajutat!

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Răspuns:

Explicație pas cu pas: