Ta có: $a+b+c=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$$\Leftrightarrow a+b+c=ab+bc+ca$$\Leftrightarrow abc-ab-bc-ca+a+b+c-1=0$$\Leftrightarrow (a-1)(b-1)(c-1)=0$$\Leftrightarrow \left[\begin{array}{l}a=1\\b=1\\c=1\end{array}\right.\Rightarrow M=0$
Ta có: $a+b+c=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$$\Leftrightarrow a+b+c=ab+bc+ca$$\Leftrightarrow abc-ab-bc-ca+a+b+c-1=0$$\Leftrightarrow (a-1)(b-1)(c-1)=0$$\Leftrightarrow \left[\begin{array}{l}a=1\\b=1\\c=1\end{array}\right.\Rightarrow M=1$
Ta có: $a+b+c=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$$\Leftrightarrow a+b+c=ab+bc+ca$$\Leftrightarrow abc-ab-bc-ca+a+b+c-1=0$$\Leftrightarrow (a-1)(b-1)(c-1)=0$$\Leftrightarrow \left[\begin{array}{l}a=1\\b=1\\c=1\end{array}\right.\Rightarrow M=
0$