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

=> a/a'=c/c'

\(\frac{a}{a'}+\frac{b'}{b}=1\)=> \(\frac{a}{a'}.\frac{b}{b'}+\frac{b'}{b}.\frac{b}{b'}=\frac{b}{b'}\)=> \(\frac{ab}{a'b'}+1=\frac{b}{b'}=1-\frac{c'}{c}\)

=> \(\frac{ab}{a'b'}=-\frac{c'}{c}\)=> abc = - a'b'c' => abc + a'b'c' = 0

chua hoc phan nay nen cug cha bt giai luon

a/a' + b'/b = 1 <=> ab + a'b' = a'b <=> abc + a'b'c = a'bc (1) (vì c # 0)
b/b' + c'/c = 1 <=> bc + b'c' = b'c <=> a'bc + a'b'c' = a'b'c (2) (vì a' # 0)
(1) + (2) => đpcm

\(\dfrac{a}{a'}+\dfrac{b'}{b}=1\Rightarrow\dfrac{a}{a'}\cdot\dfrac{b}{b'}+\dfrac{b'}{b}\cdot\dfrac{b}{b'}=\dfrac{b}{b'}\Rightarrow\dfrac{ab}{a'b'}+1=\dfrac{b}{b'}\left(1\right)\)

\(\dfrac{b}{b'}+\dfrac{c'}{c}=1\Rightarrow\dfrac{b}{b'}=1-\dfrac{c'}{c}\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\Rightarrow\dfrac{ab}{a'b'}=-\dfrac{c'}{c}\Rightarrow abc=-a'b'c'\Rightarrow abc+a'b'c'=0\)

Vậy \(abc+a'b'c'=0\left(dpcm\right)\)

Help me vs mấy chế ơi