Ta có : \(\frac{x}{y}=\frac{2}{3}\Rightarrow x=\frac{2}{3}y\)
Thay \(x=\frac{2}{3}y\)vào A , ta được :
\(A=\frac{5.\frac{2}{3}y+3y}{6.\frac{2}{3}y-7y}\)
\(\Rightarrow A=\frac{\frac{10}{3}y+3y}{4y-7y}\)
\(\Rightarrow A=\frac{\left(\frac{10}{3}+3\right)y}{-3y}\)
\(\Rightarrow A=\frac{\frac{19}{3}y}{-3y}\)
\(\Rightarrow A=\frac{\frac{19}{3}}{-3}\)
\(\Rightarrow A=\frac{19}{3}.-\frac{1}{3}\)
\(\Rightarrow A=-\frac{19}{9}\)
Vậy \(A=-\frac{19}{9}\)
\(\frac{x}{y}\)= \(\frac{2}{3}\)=>\(\frac{x}{2}\)=
\(\frac{x}{2}\)\(=\frac{y}{3}\)=>\(\frac{5x}{10}\)\(=\frac{3y}{9}\)\(=\frac{5x+3y}{10+9}\)\(=\frac{5x+3y}{19}\)\(=\frac{6x}{12}\)\(=\frac{7y}{21}\)\(=\frac{6x-7y}{12-21}\)\(=\frac{6x-7y}{-9}\)
\(=>\frac{5x+3y}{6x-7y}\)\(=\frac{-19}{9}\)
\(#Hoktot~\)