\(3^x.\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}\right)=39\)
\(3^x.\frac{13}{27}=39\)
\(3^x=\frac{39.27}{13}=3.27=3.3^3=3^4\)
\(\Rightarrow x=4\)
\(\frac{3x-1}{2}\)=\(\frac{2y-3}{5}\)=\(\frac{3x-2y+2}{6x}\) .Tìm x và y
\(3^{x-1}+3^{x-2}+3^{x-3}=39\)
\(\Leftrightarrow3^x:3+3^x:3^2+3^x:3^3=39\)
\(\Leftrightarrow3^x.\frac{1}{3}+3^x.\frac{1}{3^2}+3^x.\frac{1}{3^3}=39\)
\(3^x.\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}\right)=39\)
\(3^x.\frac{13}{27}=39\)
\(3^x=39:\frac{13}{27}\)
\(3^x=3^4\)
\(\Leftrightarrow x=4\)
