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]
Atunci vom avea
[tex]n^{-\sqrt[n]{\frac{1}{n^{2} } } } = n^{1}, evident \:ca[/tex] ⇒ [tex]{-\sqrt[n]{\frac{1}{n^{2} } } } = 1, deci \: atunci \: n= 1\: sau \: -1[/tex]
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!
