2 câu trả lời
Theo đề bài, ta có:
`a/b=c/d`
`→a/b-2=c/d-2`
`→a/b-2/b=c/d-2/d`
`→{a-2}/b={c-2}/d`
Đặt $\dfrac{a}{b}=\dfrac{c}{d}=k(k\ne 0)$
$=>a=bk,c=dk\\\dfrac{a-2b}{b}=\dfrac{bk-2b}{b}=k-2\\\dfrac{c-2d}{d}=\dfrac{dk-2d}{d}=k-2\\=> \dfrac{a-2b}{b}=\dfrac{c-2d}{d}$