Ta có:
\(\dfrac{7a-11b}{4a+5b}=\dfrac{7c-11d}{4c+5d}\)
\(\Rightarrow\dfrac{7a-11b}{7c-11d}=\dfrac{4a+5b}{4c+5d}\)
\(\Leftrightarrow\dfrac{7a}{7c}=\dfrac{11b}{11d}=\dfrac{4a}{4c}=\dfrac{5b}{5d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Mặt khác:
\(\dfrac{a}{c}=\dfrac{b}{d}\Leftrightarrow\dfrac{a}{b}=\dfrac{c}{d}\left(đpcm\right)\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}\)= k
Vì \(\dfrac{a}{b}=k\) = > a = bk
Vì \(\dfrac{c}{d}=k\) = > c = dk
Ta có: \(\dfrac{7a-11b}{4a+5b}=\dfrac{7.bk-11b}{4.bk+5b}=\dfrac{\left(7.11\right).b.\left(k-1\right)}{\left(4.5\right).b.\left(k+1\right)}\dfrac{\left(7.11\right).\left(k-1\right)}{\left(4.5\right).\left(k+1\right)}\)(1)
\(\dfrac{7c-11d}{4c+5d}=\dfrac{7.dk-11d}{4.dk+5d}=\dfrac{\left(7.11\right).d.\left(k-1\right)}{\left(4.5\right).d.\left(k+1\right)}=\dfrac{\left(7.11\right).\left(k-1\right)}{\left(4.5\right).\left(k+1\right)}\left(2\right)\)Từ (1) và (2) = > \(\dfrac{7a-11b}{4a+5b}=\dfrac{7c-11d}{4c+5d}\)
