Réponse :
Bonjour,
Explications étape par étape
1)
[tex]a,b \not=0\\\\\dfrac{a}{b} =\dfrac{b}{a+b} \\\\\dfrac{b}{a} =\dfrac{a+b}{b} \\\\\dfrac{b}{a} =1+\dfrac{a}{b} \\\\(\dfrac{b}{a})^2 =\dfrac{b}{a}(1+\dfrac{a}{b}) \\\\(\dfrac{b}{a})^2 =\dfrac{b}{a}+1 \\[/tex]
2)
[tex]Soit \ x=\dfrac{b}{a} \\\\(\dfrac{b}{a})^2 =\dfrac{b}{a}+1 \Longrightarrow\ x^2=x+1\\[/tex]
3)
[tex]x^2-x-1=0\\\Delta=1+4=5\\\phi=\dfrac{1+\sqrt{5} }{2} \\\\\overline{\phi}=\dfrac{1-\sqrt{5} }{2} \\[/tex]
3)
[tex]-\dfrac{1}{\phi} =-\dfrac{2}{1+\sqrt{5} } \\=-\dfrac{2*(1-\sqrt{5})}{(1+\sqrt{5})*(1-\sqrt{5}) } \\\\=\dfrac{-2*(1-\sqrt{5})}{-4} \\\\=\dfrac{1-\sqrt{5} }{2} =\overline{\phi}\\[/tex]
4)
[tex]x\not=0,\ x^2=1+x\Longrightarrow\ x=\sqrt{1+x} \\\\donc\ \phi=\sqrt{1+\phi} ....\\[/tex]
évident
5)
[tex]x^2=1+x\ \Longrightarrow\ x=\dfrac{1}{x} +1\\\\donc\ \phi=1+\dfrac{1}{\phi} ....\\[/tex]
évident
5)
[tex]X=\sqrt{2+X} \\\\\Longrightarrow\ X^2-X-2=0\\\\\Delta=1+8=3^2\\\\X=\dfrac{1+3}{2} =2\ ou\ X=\dfrac{1-3}{2} =-1\\[/tex]
