Vì\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\)
Đặt \(\frac{a}{c}=\frac{b}{d}\) = k
=> a = ck , b = dk
Thay a = ck , b = dk vào \(\frac{7a-11b}{4a+5b}\)ta có :
\(\frac{7a-11b}{4a+5b}=\frac{7.ck-11dk}{4ck+5dk}=\frac{k\left(7c-11d\right)}{k\left(4c+5d\right)}=\frac{7c-11d}{4c+5d}\)
Vậy \(\frac{7a-11b}{4a+5b}=\frac{7c-11d}{4c+5d}\)