Răspuns :
[tex]11^{22} ~~~~ 22^{11}\\
11^{11}\cdot 11^{11}>2^{11}\cdot 11^{11}\\
\\
15^{90}~~~~~~~6^{135}\\
(15^2)^{45}~~~~~~ (6^3)^{45}\\
225^{45} \ \textgreater \ 216^{45}
[/tex]
11^22 =11^11 x 11^11
22^11 = 2^11 x11^11
=> 11^11 x11^11 > 2^11x11^11
11^22 > 22^11
15^90 = 3^90 x 5^90
6^135 = 2^135 x 3^135=
Impartim prin 3^90 => 5^90 fata de 2^135 x 3^45
5^90 = (5^2)^45 = 25^45
2^135 x 3^45 = (2^45 x3^45) x 2^90 = 6^45 x (2^2)^45 =6^45 x 4^45 =24^45
Deci 15^90 > 6^135 deoarece 25^45 > 24^45
22^11 = 2^11 x11^11
=> 11^11 x11^11 > 2^11x11^11
11^22 > 22^11
15^90 = 3^90 x 5^90
6^135 = 2^135 x 3^135=
Impartim prin 3^90 => 5^90 fata de 2^135 x 3^45
5^90 = (5^2)^45 = 25^45
2^135 x 3^45 = (2^45 x3^45) x 2^90 = 6^45 x (2^2)^45 =6^45 x 4^45 =24^45
Deci 15^90 > 6^135 deoarece 25^45 > 24^45
