Question

\(Cho:\frac{a}{b}=\frac{c}{d}.CMR:\frac{2017\times a-b}{a}=\frac{2017\times c-d}{c}\)

Answer 1

Ta có : \(\frac{a}{b}=\frac{c}{d}\)

Suy ra : \(\frac{a}{c}=\frac{b}{d}=\frac{2017a}{2017c}=\frac{2017a-b}{2017c-d}\)

Nên : \(\frac{a}{c}=\frac{2017a-b}{2017c-d}\)

Do đó : \(\frac{2017a-b}{a}=\frac{2017c-d}{c}\) (đpcm)

Answer 2

\(\frac{a}{b}=\frac{c}{d}\) \(\Rightarrow\frac{a}{c}=\frac{b}{d}\)

\(\Rightarrow\frac{2017a}{2017c}=\frac{b}{d}\)

Áp dụng tính chất dãy tỉ số bằng nhau ,ta có :

\(\frac{2017a}{2017c}=\frac{b}{d}=\frac{2017a-b}{2017c-d}\)

\(\Rightarrow\frac{2017a-b}{2017c-d}=\frac{b}{d}=\frac{a}{c}\)

\(\Rightarrow\frac{2017a-b}{2017c-d}=\frac{a}{c}\)

\(\Rightarrow\frac{2017a-b}{a}=\frac{2017c-d}{c}\)

Answer 3

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)

Ta có:\(\frac{2017a-b}{a}=\frac{2017bk-b}{bk}=\frac{b\left(2017k-1\right)}{bk}=\frac{2017k-1}{k}\left(1\right)\)

           \(\frac{2017c-d}{c}=\frac{2017dk-d}{dk}=\frac{d\left(2017k-1\right)}{dk}=\frac{2017k-1}{k}\left(2\right)\)

                     Từ (1) và (2) suy ra:\(\frac{2017a-b}{a}=\frac{2017c-d}{c}\)

Answer 4

Ta có : ab =cd 

Suy ra : ac =bd =2017a2017c =2017a−b2017c−d 

Nên : ac =2017a−b2017c−d 

Do đó : 2017a−ba =2017c−dc  (đpcm)