Question

Cho \(b\ne-d;b\ne-3d;b\ne0;d\ne0\) và \(\dfrac{a+3c}{b+3d}=\dfrac{a+c}{b+d}\) . Chứng minh : \(\dfrac{a}{b}=\dfrac{c}{d}\)

Answer 1

Ta có: \(\dfrac{a+3c}{b+3d}=\dfrac{a+c}{b+d}\left(b\ne-d;b\ne-3d;b\ne0;d\ne0\right)\)

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

+, \(\dfrac{a+3c}{b+3d}=\dfrac{a+c}{b+d}=\dfrac{a+3c-\left(a+c\right)}{b+3d-\left(b+d\right)}=\dfrac{a+3c-a-c}{b+3d-b-d}=\dfrac{2c}{2d}=\dfrac{c}{d}\)

Khi đó: \(\dfrac{a+c}{b+d}=\dfrac{c}{d}\)

+, \(\dfrac{a+c}{b+d}=\dfrac{c}{d}=\dfrac{a+c-c}{b+d-d}=\dfrac{a}{b}\) (đpcm)

Answer 2

Áp dụng t/c dãy tỉ số bằng nhau:

\(\dfrac{a+3c}{b+3d}=\dfrac{a+c}{b+d}=\dfrac{a+3c-\left(a+c\right)}{b+3d-\left(b+d\right)}=\dfrac{2c}{2d}=\dfrac{c}{d}\) (1)

\(\dfrac{a+3c}{b+3d}=\dfrac{a+c}{b+d}=\dfrac{3a+3c}{3b+3d}=\dfrac{a+3c-\left(3a+3c\right)}{b+3d-\left(3b+3d\right)}=\dfrac{-2a}{-2b}=\dfrac{a}{b}\) (2)

(1);(2) \(\Rightarrow\dfrac{a}{b}=\dfrac{c}{d}\)