Đề \(\Rightarrow\frac{a+b}{ab}=\frac{b+c}{bc}\)\(\Leftrightarrow\frac{a}{ab}+\frac{b}{ab}=\frac{b}{bc}+\frac{c}{bc}\)\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{b}+\frac{1}{c}\)

\(\Leftrightarrow\frac{1}{a}=\frac{1}{c}\Rightarrow a=c\Leftrightarrow ab=bc\)

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

Đề sai hả bạn ?

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

\(\Leftrightarrow\frac{10a+b}{a+b}=\frac{10b+c}{b+c}\)

\(\Leftrightarrow\frac{a+b+9a}{a+b}=\frac{b+c+9b}{b+c}\)

\(\Leftrightarrow1+\frac{9a}{a+b}=1+\frac{9b}{b+c}\)

\(\Leftrightarrow\frac{9a}{a+b}=\frac{9b}{b+c}\)

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

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

\(\Leftrightarrow ab+ac=ab+b^2\)

\(\Leftrightarrow ac=b^2\)

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

Ta có:

\(\frac{\overline{ab}}{a+b}=\frac{\overline{bc}}{b+c}\Rightarrow\frac{\overline{ab}}{\overline{bc}}=\frac{a+b}{b+c}=\frac{\overline{ab}-\left(a+b\right)}{\overline{bc}-\left(b+c\right)}\)

\(=\frac{10a+b-a-b}{10b+c-b-c}=\frac{9a}{9b}=\frac{b}{a}\)

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

Vậy: \(\frac{a}{b}=\frac{b}{c}\left(b,c\ne0\right)\)

Bn ơi mk nghĩ đề phải là : giả thuyết \(c\ne0\)bn nhé.......

#kiseki no enzeru#

hok tốt

Đặt \(\frac{a}{b}=\frac{c}{d}=k\) thì \(a=bk,c=dk\).

\(\frac{2a+3b}{2a-3b}=\frac{2bk+3b}{2bk-3b}=\frac{b\left(2k+3\right)}{b\left(2k-3\right)}=\frac{2k+3}{2k-3}\\ \frac{2c+3d}{2c-3d}=\frac{2dk+3d}{2dk-3d}=\frac{d\left(2k+3\right)}{d\left(2k-3\right)}=\frac{2k+3}{2k-3}\)

Do đó: \(\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\)

\(\frac{\overline{ab}}{a+b}=\frac{\overline{bc}}{b+c}=\frac{10a+b}{a+b}=\frac{10b+c}{b+c}\)

áp dụng t/c dãy tỉ số bằng nhau ta có:

\(\frac{\overline{ab}}{a+b}=\frac{\overline{bc}}{b+c}=\frac{10a+b}{a+b}=\frac{10b+c}{b+c}=\frac{10a+11b+c}{a+2b+c}\)

\(\Rightarrow\frac{10a+b}{a+b}=\frac{10a+11b+c}{a+2b+c}\Rightarrow\left(10a+b\right).\left(a+2b+c\right)=\left(a+b\right).\left(10a+11b+c\right)\)

\(10a^2+20ab+10ac+ab+2b^2+bc=10a^2+11ab+ac+10ab+11b^2+bc\)

\(\Rightarrow9ac=9b^2\Rightarrow ac=b^2\Rightarrow\frac{a}{b}=\frac{b}{c}\left(đpcm\right)\)

p/s: bài này khó chơi lém, đoạn mk giản đơn hai vế ko hiểu ib vs mk :))

a)\(\frac{ab}{cd}=\frac{bk.b}{dk.b}=\frac{b^2}{d^2}\left(1\right)\)

\(\frac{a^2-b^2}{c^2-d^2}=\frac{b^2k^2-b^2}{d^2k^2-d^2}=\frac{b^2\left(k^2-1\right)}{d^2\left(k^2-1\right)}=\frac{b^2}{d^2}\left(2\right)\)

từ\(\left(1\right)\)và\(\left(2\right)\)\(\Rightarrow\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)

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Xem ở lick này nhé (mình gửi cho)

Học tốt!!!!!!!!!!!!!

@@ chị linh Link dài vậy giải lun phải hơn không

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

\(\Rightarrow\frac{2a}{2c}=\frac{3b}{3d}\)

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

\(\frac{2a}{2c}=\frac{3b}{3d}=\frac{2a-3b}{2c-3d}=\frac{2a+3b}{2c+3d}\)

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