Question

Chứng minh rằng nếu \(\dfrac{a}{b}=\dfrac{c}{d}\ne1\)(a,b,c,d\(\ne\)0)Thì \(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)

Answer 1

Ta có : \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)

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

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

Tương tự :

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

Từ \(\left(1\right)\) và \(\left(2\right)\) suy ra : \(\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\Rightarrow\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)

Vậy \(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\) khi \(\dfrac{a}{b}=\dfrac{c}{d}\)

Answer 2

Đặt: a/b = c/d = k => a = bk, c = dk
Ta có:
a + b/a - b = bk + b/bk - b = b(k+1)/ b(k-1) = k+1/k-1 (1)
c + d/c- d = dk +d/ dk - d = d(k+1)/d(k-1) = k+1/k-1 (2)
Từ (1) và (2) => a+b/a-b = c+d/c-d

Answer 3

\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow a=bk;c=dk\)

\(\Rightarrow\dfrac{a+b}{a-b}=\dfrac{bk+b}{bk-b}=\dfrac{b\left(k+1\right)}{b\left(k-1\right)}=\dfrac{k+1}{k-1}\)

\(\Rightarrow\dfrac{c+d}{c-d}=\dfrac{dk+d}{dk-d}=\dfrac{d\left(k+1\right)}{d\left(d-1\right)}=\dfrac{k+1}{k-1}\)

\(\dfrac{k+1}{k-1}=\dfrac{k+1}{k-1}\)

\(\Rightarrow\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\rightarrowđpcm\)

Answer 4

Ta có : \(\dfrac{a}{b}=\dfrac{c}{d}\) \(\Rightarrow\)\(\dfrac{a}{c}=\dfrac{b}{d}\)

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

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