Bài này có 2 cách nè:
Cách 1:
\(\frac{a+b}{c+d}=\frac{a-2b}{c-2d}=\frac{\left(a+b\right)-\left(a-2b\right)}{\left(c+d\right)-\left(c-2d\right)}=\frac{a+b-a+2b}{c+d-c+2d}=\frac{3b}{3d}=\frac{b}{d}\left(1\right)\)
\(\frac{a+b}{c+a}=\frac{a-2b}{c-d}=\frac{2a+2b}{2c+2d}=\frac{2-ab+2a+2b}{c-2d+2c+2d}=\frac{3a}{3c}=\frac{a}{c}\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\) \(\Rightarrow\frac{b}{d}=\frac{a}{c}\) Suy ra: \(\frac{a}{b}=\frac{c}{d}\)
Cách 2:
\(\frac{a+b}{c+d}=\frac{a-2b}{c-2d}\)
\(\Rightarrow\left(a+b\right).\left(c-2d\right)=\left(a-2b\right).\left(c+d\right)\)
\(\Rightarrow a.c-2a.d+b.c-2b.d=a.c+a.d-2b.c-2b.d\)
\(\Rightarrow a.c-a.c-2a.d-a.d+b.c-2b.c-2b.d+2b.d=0\)
\(\Rightarrow-3a.d+3b.d=0\)
\(\Rightarrow3b.c=3a.d\)
\(\Rightarrow b.c=a.d\)
\(\Rightarrow\frac{a}{b}=\frac{c}{d}\)
