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