Rozwiąż równanie: 1. [latex]9 ^{ sqrt{x} } -8*3^ {sqrt{3}} -9=0[/latex] 2. [latex](3-2 sqrt{2} ) ^{x} +(3+2 sqrt{2} )^{x}=2[/latex] do 2 wskazówka : [latex]2+ sqrt{3} = frac{1}{2- sqrt{3} } [/latex]

[latex]9^{sqrt{x}}-8 cdot 3^{sqrt{3}}-9=0 \ 9^{sqrt{x}}=9+8 cdot 3^{sqrt{3}} \ sqrt{x}=log_{9}(9+ 8 cdot 3^{sqrt{3}}) quad /(...)^{2} \ x=log^{2}_{9}(9+8 cdot 3^{sqrt{3}})[/latex] W drugim zauważasz że  [latex]3-2sqrt{2}=frac{1}{3+2sqrt{2}} \ hbox{Dokonujesz podstawienia:} t=(3+2sqrt{2})^{x} \ frac{1}{t}+t=2 \ frac{1+t^{2}}{t}=2 \ 1+t^{2}=2t \ t^{2}-2t+1=0 \ (t-1)^{2}=0 \ t=1 \ hbox{Skad:} \ (3+2sqrt{2})^{x}=1 \ (3+2sqrt{2})^{x}=(3+2sqrt{2})^{0} \ x=0[/latex]