Question

Cho \(\frac{a+b}{c+d}=\frac{a-2b}{c-2d}\)

CMR: \(\frac{a}{b}=\frac{c}{d}\)

Answer 1

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

=> \(\frac{a}{b}=\frac{k.c}{k.d}=\frac{c}{d}\left(đpcm\right)\)

Answer 2

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

=> \(\frac{a}{b}=\frac{c}{d}\left(đpcm\right)\)