Ta có : \(a^2=bc\Rightarrow\frac{a}{c}=\frac{b}{a}\)

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

    \(\frac{a}{c}=\frac{b}{a}=\frac{a+b}{c+a}=\frac{a-b}{c-a}\)

Từ \(\frac{a+b}{c+a}=\frac{a-b}{c-a}\Leftrightarrow\frac{a+b}{a-b}=\frac{c+a}{c-a}\)

Vậy nếu \(a^2=bc\Rightarrow\frac{a+b}{a-b}=\frac{c+a}{c-a}\)

Ta có: \(a^2=bc\Rightarrow\frac{a}{c}=\frac{b}{a}\)

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

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

Từ \(\frac{a+b}{a-b}=\frac{a-b}{c-a}\Rightarrow\frac{a+b}{a-b}=\frac{c+a}{c-a}\)

Vậy: \(\frac{a+b}{a-b}=\frac{c+a}{c-a}\)

Cảm Ơn các bạn nhé!!!!!!!!!@

Đặt\(\frac{a}{b}=\frac{c}{d}=k\)

\(\Rightarrow a=bk ;c=dk\)

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

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

Từ (1) và (2)=> \(\frac{a-b}{a}=\frac{c-d}{c}\)

\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)

\(\frac{1}{c}=\frac{1}{2}\left(\frac{a+b}{ab}\right)\)

\(\Rightarrow2ab=c\left(a+b\right)\)

\(\Rightarrow ab+ab=ca+bc\)

\(\Rightarrow ab-cb=ac-ab\)

\(\Rightarrow b\left(a-c\right)=a\left(c-b\right)\)

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

Trả lời :........................................................

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

Hk tốt,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,

Học sinh giỏi 6A

ta có: \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)

\(=\frac{1}{c}\times2=\frac{1}{a}+\frac{1}{b}\)

\(=\frac{2}{c}=\frac{1}{a}+\frac{1}{b}\)

\(=\frac{2}{c}=\frac{b+a}{ab}\)

= \(c\left(b+a\right)=ab\times2\)

= cb +ca = ab+ab

= ab - cb = ac-ab

\(=b\left(a-c\right)=a\left(c-b\right)\)

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

\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)

\(\frac{1}{c}=\frac{1}{2a}+\frac{1}{2b}\)

\(\frac{1}{c}=\frac{a+b}{2ab}\)

\(2ab=c\left(a+b\right)\)

\(ab+ab=ac+bc\)

\(ab-bc=ac-ab\)

\(b\left(a-c\right)=a\left(c-b\right)\)

\(\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}\left(đpcm\right)\)

giúp mình bài này với

so sánh bằng cách nhanh nhất

a 2013 phần 2012 và 13 phần 12

b 15 phần 46 và 21 phần 62

Từ \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\Rightarrow\frac{1}{c}=\frac{1}{2}\left(\frac{a+b}{ab}\right)\)

\(\Rightarrow\frac{1}{c}=\frac{a+b}{2ab}\)

\(\Rightarrow2ab=c.\left(a+b\right)\)

\(\Rightarrow ab+ab=ac+bc\)

\(\Rightarrow ab-bc=ac-ab\)

\(\Rightarrow b.\left(a-c\right)=a.\left(c-b\right)\)

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

\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)

=>\(\frac{1}{c}=\frac{a+b}{2ab}\)

=> 2ab = c(a+b)

=> ab+ab = ac+bc

=> ab - bc = ac - ab

=> b(a-c) = a(c-b)

=> \(\frac{a}{b}=\frac{a-b}{c-b}\left(đpcm\right)\)