Răspuns:
Explicație pas cu pas:
[tex]a= 5^{3005} - 4*5^{3004} - 5^{3003} = 5^2*5^{3003} - 4*5*5^{3003} - 5^{3003} = 5^{3003}(5^2 - 4*5-1) = 5^{3003}(25-20-1)=4*5^{3003}[/tex]
[tex]b= 2^{7011}-2^{7010} - 2^{7009} = 2^2*2^{7009}-2*2^{7009}-2^{7009} = 2^{7009}(2^2-2-1) = 2^{7009}(4-2-1)=2^{7009}[/tex]
[tex]c= 3^{5008} -2*3^{5007}-3^{5006}-2*3^{5005} = 3^3*3^{5005}-2*3^2*3^{5005}-3*3*3^{5005} -2*3^{5005} = 3^{5005}(3^3-2*3^2-3-2) =3^{5005}(27-18-3-2)=4*3^{5005}[/tex]
Deci:
[tex]a=4*5^{3003}=4*(5^3)^{1001} = 4*125^{1001}\\b=2^{7009} = 2^2*2^{7007}=4*(2^7)^{1001}=4*128^{1001}\\c= 4*3^{5005}=4*(3^5)^{1001} = 4*243^{1001}\\[/tex]
asadar in ordine crescatoare se scriu a, b, c
b) folosim trei puteri ale lui 2: [tex]2^0 , 2^2 , 2^7:[/tex]
[tex]2017=2^{7}*2^{2}*2^{2} - \frac{2^7}{2^2} + 2^0[/tex]
Intadevar:
[tex]128*4*4 - \frac{128}{4} + 1 = 2048 - 32 + 1 = 2017[/tex]
log4 + 5005*log3
