a: Đặt a/b=c/d=k

=>a=bk; c=dk

\(\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{k}{k-1}\)

\(\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{k}{k-1}\)

Do đó: \(\dfrac{a}{a-b}=\dfrac{c}{c-d}\)

b: Đặt a/b=c/d=k

=>a=bk; c=dk

\(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bk+b}{dk+d}\right)^2=\dfrac{b^2}{d^2}\)

\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{b^2k^2+b^2}{d^2k^2+d^2}=\dfrac{b^2}{d^2}\)

DO đó: \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)

Đặt \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}=\dfrac{e}{f}=\dfrac{a+b+c+d+e}{b+c+d+e+f}=k\)

Ta có:

\(\dfrac{a}{f}=\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}.\dfrac{d}{e}.\dfrac{e}{f}=k^5=\left(\dfrac{a+b+c+d+e}{b+c+d+e+f}\right)^5\)

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

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

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

Từ 2 điều trên suy ra \(\dfrac{a+b}{b}=\dfrac{c+d}{d}\)

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Giải:

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 có:

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

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

Vậy...

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

=>\(\left\{{}\begin{matrix}a=b.k\\c=d.k\end{matrix}\right.\) (1)

Thay (1) vào:

\(\dfrac{a+b}{a-b}=\dfrac{b.k+b}{b.k-b}=\dfrac{b.\left(k+1\right)}{b.\left(k-1\right)}=\dfrac{k+1}{k-1}\) (2)

\(\dfrac{c+d}{c-d}=\dfrac{d.k+d}{d.k-d}=\dfrac{d.\left(k+1\right)}{d.\left(k-1\right)}=\dfrac{k+1}{k-1}\) (3)

Từ (2) và (3) =>\(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}=\dfrac{k+1}{k-1}\)

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

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

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

Áp dụng tính chất hoán vị ta được:

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

Vậy từ tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)

a: \(\dfrac{a}{b}=\dfrac{c}{d}\)

\(\Leftrightarrow\dfrac{a}{b}-1=\dfrac{c}{d}-1\)

hay \(\dfrac{a-b}{b}=\dfrac{c-d}{d}\)

Ta có :

\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\)

\(\Leftrightarrow\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}\)\(\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{ab}{cd}\)

\(\Leftrightarrow\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{ab}{cd}\)

\(\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}\)

\(\Leftrightarrow\dfrac{a}{b}=\dfrac{c}{d}\left(đpcm\right)\)

(a² + b²) / (c² + d²) = ab/cd
<=> (a² + b²)cd = ab(c² + d²)
<=> a²cd + b²cd = abc² + abd²
<=> a²cd - abc² - abd² + b²cd = 0
<=> ac(ad - bc) - bd(ad - bc) = 0
<=> (ac - bd)(ad - bc) = 0
<=> ac - bd = 0 hoặc ad - bc = 0
<=> ac = bd hoặc ad = bc
<=> a/b = d/c hoặc a/b = c/d (đpcm)