Rozwiązanie:
Równanie:
[tex](\sqrt{3-2\sqrt{2} } )^{x}+(\sqrt{3+2\sqrt{2} } )^{x}=6[/tex]
Zauważmy, że:
[tex]$\frac{1}{\sqrt{3-2\sqrt{2} } } =\frac{\sqrt{3+2\sqrt{2} } }{\sqrt{9-8} } =\sqrt{3+2\sqrt{2} }[/tex]
Zatem mamy równanie:
[tex]$(\sqrt{3-2\sqrt{2} } )^{x}+(\frac{1}{\sqrt{3-2\sqrt{2} } } )^{x}=6[/tex]
Podstawmy [tex]t=(\sqrt{3-2\sqrt{2} } )^{x}[/tex], gdzie [tex]t>0[/tex] :
[tex]$ t+\frac{1}{t} =6[/tex]
[tex]t^{2}-6t+1=0[/tex]
[tex]\Delta=36-4 \cdot 1 \cdot 1=32\\[/tex]
[tex]$t_{1}=\frac{6-4\sqrt{2} }{2} =3-2\sqrt{2}[/tex]
[tex]$t_{2}=\frac{6+4\sqrt{2} }{2} =3+2\sqrt{2}[/tex]
Zatem:
[tex](\sqrt{3-2\sqrt{2} } )^{x}=3-2\sqrt{2} \vee (\sqrt{3-2\sqrt{2} } )^{x}=3+2\sqrt{2}[/tex]
[tex]x=2 \vee x=-2[/tex]
