Ta có :\(\dfrac{a}{a^,}+\dfrac{b^,}{b}=1\Rightarrow\left(\dfrac{a}{a^,}+\dfrac{b^,}{b}\right).\dfrac{b}{b^,}=\dfrac{b}{b^,}\Rightarrow\dfrac{ab}{a^,b}+1=\dfrac{b}{b^,}=1-\dfrac{c^,}{c}\)( vì \(\dfrac{b}{b^,}+\dfrac{c^,}{c}=1\))
\(\Rightarrow\dfrac{ab}{a^,b^,}=-\dfrac{c^,}{c}\Rightarrow abc=-a^,b^,c^,\Rightarrow abc+a^,b^,c^,=0\)( đpcm )