Đặt \(\frac{x}{2}=\frac{2y}{5}=\frac{3z}{7}=k\)
\(\Rightarrow\hept{\begin{cases}x=2k\\y=\frac{5}{2}k\\z=\frac{7k}{3}\end{cases}}\)
Thay vô rồi tính tiếp nhé!
bạn có thể nói rõ hơn ko?
Mk chưa hiểu lắm
Đặt \(\frac{x}{2}=\frac{2y}{5}=\frac{3z}{7}=k\)
\(\Rightarrow\hept{\begin{cases}x=2k\\y=\frac{5k}{2}\\z=\frac{7k}{3}\end{cases}}\)
\(\Rightarrow A=\frac{2.2k+5.\frac{5k}{2}-3.\frac{7k}{3}}{7.2k+\frac{5k}{2}-5.\frac{7k}{3}}\)
\(=\frac{-3k+\frac{25k}{2}}{14k-\frac{55k}{6}}\)
\(=\frac{19k.6}{2.29k}=\frac{57}{29}\)
Ý là vậy :P
Bài giải
\(\frac{x}{2}=\frac{2y}{5}=\frac{3z}{7}\text{ }\Rightarrow\hept{\begin{cases}y=\frac{5x}{4}\\z=\frac{7x}{6}\end{cases}}\)
\(A=\frac{2x+5y-3z}{7x+y-5z}=\frac{2x+\frac{25x}{4}-\frac{7x}{2}}{7x+\frac{5x}{4}-\frac{35x}{6}}=\frac{x\left(2+\frac{25}{4}-\frac{7}{2}\right)}{x\left(7+\frac{5}{4}-\frac{35}{6}\right)}=\frac{\frac{19}{4}}{\frac{29}{12}}=\frac{19}{4}\cdot\frac{12}{29}=\frac{57}{29}\)
Vậy \(A=\frac{57}{29}\)
!!!!!!!!!!!!!!!!!!!!!!!
