Đặt \(\frac{a+b}{c+d}=\frac{a-2b}{c-2d}=k\)
=> \(\begin{cases}a+b=k.\left(c+d\right)=k.c+k.d\\a-2b-k.\left(c-2d\right)=k.c-k.2d\end{cases}\)
=> (a + b) - (a - 2b) = (k.c + k.d) - (k.c - k.2d)
=> a + b - a + 2b - k.c + k.d - k.c + k.2d
=> 3b = 3kd
=> b = kd
Mà a + b = k.c + k.d
=> a = k.c
=> \(\frac{a}{b}=\frac{k.c}{k.d}=\frac{c}{d}\left(đpcm\right)\)
Cách 2:
Ta có: \(\frac{a+b}{c+d}=\frac{a-2b}{c-2d}\)
=> (a + b).(c - 2d) = (c + d).(a - 2b)
=> (a + b).c - (a + b).2d = (c + d).a - (c + d).2b
=> ac + bc - 2ad - 2bd = ac + ad - 2bc - 2bd
=> ac + bc - 2ad - 2bd - ac - ad + 2bc + 2bd = 0
=> 3bc - 3ad = 0
=> 3bc = 3ad
=> bc = ad
=> \(\frac{a}{b}=\frac{c}{d}\left(đpcm\right)\)
