Radodanaask
a fost răspuns

Ajutorrr!! imi trebuie neaparat !! Este o urgenta !!
      2   4   6       2n              2   3       2n+2
(1+3 +3 +3 +...+3  ) * x=3 +3 +3 +..+ 3
 
          2   3         2x+1        4 507             811
2 +2 +2 +2 +...+2     +32*((2 )      )  =72*32

Răspuns :

Danaradu70ask
Fie 3^2=9 ⇒ 9^2=(3^2)^2=3^4, 9^3=(3^2)^3=3^6......9^n=(3^2)^n=3^2n
Relatia se scrie
(9^0+9^1+....9^n)*x= 3^0+3^1+3^2+....+3^(2n+2) - 1
Folosim formula x^0+x^1+....+x^n=x^(n+1)-1/(x-1)
{[9^(n+1)-1]/(9-1)}*x=[3^(2n+3)-1]/2-1
3*(2n+3)=3^(1+2n+2)=3*3^(2n+2) 
Folosesti formula de la puteri x^(a+b)=x^a*x^b unde x=3, a=1 si b=2n+2
{[9^(n+1)-1]/8}*x=[3*3^2(n+1)-1-2]/2 (ai adus la acelasi numitor in partea dreapta)
Aducem la acelasi numitor adica inmultim in partea dreapta cu 4 si renuntam la numitor
[9^(n+1)-1]*x=3*4*[3^2]^(n+1) - 3*4
[9^(n+1)-1]*x=12*9^(n+1)-12=12[9^(n+1)-1]⇒ x=12

Sorina2007receaask
[tex](9^0+9^1+....9^n)*x= 3^0+3^1+3^2+....+3^(2n+2) - 1[/tex]
Folosim formula
[tex]x^0+x^1+....+x^n=x^(n+1)-1/(x-1) [/tex]
[tex]{[9^(n+1)-1]/(9-1)}*x=[3^(2n+3)-1]/2-1[/tex]
[tex]3*(2n+3)=3^(1+2n+2)=3*3^(2n+2)[/tex]
Folosesti formula de la puteri
[tex]x^(a+b)=x^a*x^b [/tex] unde x=3, a=1 si b=2n+2
[tex]{[9^(n+1)-1]/8}*x=[3*3^2(n+1)-1-2]/2 [/tex]  (ai adus la acelasi numitor in partea dreapta)
Aducem la acelasi numitor adica inmultim in partea dreapta cu 4 si renuntam la numitor
[tex][9^(n+1)-1]*x=3*4*[3^2]^(n+1) - 3*4 [/tex]
[tex][9^(n+1)-1]*x=12*9^(n+1)-12=12[9^(n+1)-1] ;deci x=12 [/tex]