Question

Cum se rezolva aceasta problema ?

Answer 1

 

$\displaystyle\\9^x-4\cdot3^x+3=0\\\\(3^2)^x-4\cdot3^x+3=0\\\\3^{2x}-4\cdot3^x+3=0\\\\(3^x)^2-4\cdot3^x+3=0\\\\\text{Substitutie: }~\boxed{3^x=y}\\\\y^2-3y-y+3=0\\\\y(y-3)-(y-3)=0\\\\(y-3)(y-1)=0\\(y-3)=0\implies\boxed{y_1=3}\\\\(y-1)=0\implies\boxed{y_2=1}\\\\\text{Ne intoarcem la substitutie.}\\\\3^x=3\implies \boxed{x_1=1}\\\\3^x=1\implies \boxed{x_2=0}\\\\\text{Raspuns corect: }~~\boxed{\bf~e)~0~si~1}$

Answer 2

Răspuns:

e)0.si.1

Explicație pas cu pas:

9^x-4×3^x+3=0 avem 9^x=(3^2)^x

(3^2)^x-4×3^x+2=0

folosim formula (a^m)^n=(a^n)^m si transformăm expresia (3^2)^x=(3^x)^2

(3^x)^2-4×3^x+3=0 avem (3^x)^2-4×3^x=t^2-4t

t^2-4t+3=0

t=3

t=1

3^x=3

3^x=1

x1=0

x2=1

deci S={0,1}

varianta corecta este e)0 si 1