\(\frac{x}{2}\) = \(\frac{2y}{5}\)= \(\frac{3z}{7}\)
=> \(\frac{2x}{4}\) = \(\frac{5y}{12,5}\) = \(\frac{3z}{7}\)
Áp dụng tc dãy tỉ số bằng nhau ta có:
\(\frac{2x}{4}\) = \(\frac{5y}{12,5}\) = \(\frac{3z}{7}\) = \(\frac{2x+5y-3z}{4+12,5-7}\)= \(\frac{2x+5y-3z}{9,5}\) (1)
\(\frac{x}{2}\) = \(\frac{2y}{5}\)= \(\frac{3z}{7}\)
=> \(\frac{7x}{14}\) = \(\frac{y}{2,5}\)= \(\frac{5z}{\frac{35}{3}}\)
Áp dụng tc dãy tỉ số bằng nhau ta có:
\(\frac{7x}{14}\) = \(\frac{y}{2,5}\)= \(\frac{5z}{\frac{35}{3}}\) = \(\frac{7x+ y -5z}{14+2,5-\frac{35}{3}}\)
= \(\frac{7x+ y -5z}{\frac{29}{6}}\) (2)
Từ (1) và (2) => \(\frac{2x+5y-3z}{9,5}\) = \(\frac{7x+ y -5z}{\frac{29}{6}}\)
=> \(\frac{2x+5y-3z}{7x+y-5z}\) = \(\frac{9,5}{\frac{29}{6}}\) =\(\frac{57}{29}\)
