Question

помогите решить 3-ий номер срочно

Answer 1

$begin{cases}3^{x+y}=27\3^x+3^y=12end{cases}\3^{x+y}=27\3^{x+y}=3^3\x+y=3\x=3-y\begin{cases}x=3-y\3^{3-y}+3^y=12end{cases}\3^{3-y}+3^y=12\3^3cdot3^{-y}+3^y=12\frac{27}{3^y}+3^y=12\frac{27+3^{2y}}{3^y}=12\27+3^{2y}=12cdot3^y\3^{2y}-12cdot3^y+27=0\3^y=t,;3^{2y}=t^2,;t textgreater 0\t^2-12t+27=0\D=144-4cdot27=36=6^2\t_{1,2}=frac{12pm6}2\t_1=1,;t_2=3\3^y=1Rightarrow y_1=0\3^y=3Rightarrow y_2=1\begin{cases}x=3\y=0end{cases}quadquadquadbegin{cases}x=0\y=3end{cases}$