Giải thích các bước giải:
Ta có :
$a+b+c=\dfrac 1a+\dfrac 1b+\dfrac 1c$
$\to a+b+c=\dfrac{ab+bc+ca}{abc}=ab+bc+ca$
$\to a+(b+c)=a(b+c)+bc$
$\to a=a(b+c)-(b+c)+\dfrac{1}{a}$
$\to (a-1)(b+c)+\dfrac{1}{a}-a=0$
$\to (a-1)(b+c)+\dfrac{1-a^2}{a}=0$
$\to (a-1)(b+c)+\dfrac{(1-a)(1+a)}{a}=0$
$\to (a-1)(b+c)-\dfrac{(a-1)(1+a)}{a}=0$
$\to (a-1)(b+c-\dfrac{a+1}{a})=0$
$\to (a-1)(b+c-\dfrac{a+abc}{a})=0$
$\to (a-1)(b+c-1-bc)=0$
$\to (a-1)(b-1)(1-c)=0$
$\to a=1$ hoặc $b=1$ hoặc $c=1$
$\to M=0$