Question

\(CM: \dfrac{a+b}{a-b} = \dfrac{c+d}{c-d} ta có: \dfrac{a}{b}=\dfrac{c}{d}\)

Answer 1

Đặt \(\frac{a}{b}\) = \(\frac{c}{d}\) = k

=> a = bk; c = dk

Ta đc:

VT = \(\frac{a+b}{a-b}\) = \(\frac{bk+b}{bk-b}\) = \(\frac{b\left(k+1\right)}{b\left(k-1\right)}\) = \(\frac{k+1}{k-1}\) (1)

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

Từ (1) và (2) suy ra VT = VP = \(\frac{k+1}{k-1}\)

Vậy \(\frac{a+b}{a-b}\) = \(\frac{c+d}{c-d}\) \(\rightarrow\) đpcm.

Answer 2

Ta co

a/b=c/d suy ra a/c=b/d

Ap dung tinh chat cua day ti so bang nhau ta co

a/c=b/d=a+b/c+d=a-b/c-d

Ta thay a-b/c-d=a+b/c+d

suy ra a+b/a-b=c+d/c-d (dpcm)