[tex] (\sqrt{2}+1)^x+ (\sqrt{2}-1)^x=6 \\ fie (\sqrt{2}+1)^x=t,t \geq 0 \\ t+ \frac{1}{t}=6 \\ t^{2}-6t+1=0 \\ [/tex]
Δ=36-4=32=(4√2)²
[tex] t_{1}= \frac{6-4 \sqrt{2} }{2} =3-2 \sqrt{2} \\ t_{2}= \frac{6+4 \sqrt{2} }{2}=3+2 \sqrt{2} \\ [/tex]
revenim la substitutie
[tex]( \sqrt{2}+1)^x=3-2 \sqrt{2};( \sqrt{2}+1)^x= ( \sqrt{2}+1) ^{-2} ;x=-2 \\ (\sqrt{2}+1)^x= 3+2\sqrt{2};( \sqrt{2}+1)^x=( \sqrt{2}+1)^2 ;x=2 [/tex]
