\(M=\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ac+c+1}\)
\(=\frac{a}{ab+a+1}+\frac{a.b}{a.\left(bc+b+1\right)}+\frac{c}{ac+c+1}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{abc+ab+a}+\frac{c}{ac+c+1}\)
Vì abc=1
\(=>M=\frac{a}{ab+a+1}+\frac{ab}{ab+a+1}+\frac{c}{ac+c+abc}=\frac{a}{ab+a+1}+\frac{ab}{ab+a+1}+\frac{c}{c\left(a+ab+1\right)}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{ab+a+1}+\frac{1}{ab+a+1}=\frac{ab+a+1}{ab+a+1}=1\)
Vậy M=1
\(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ac+c+1}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{abc+ab+a}+\frac{c}{ac+c+abc}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{1+ab+a}+\frac{1}{a+1+ab}=\frac{ab+a+1}{ab+a+1}=1\)
\(M=\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ac+c+1}\)
\(M=\frac{a}{ab+a+abc}+\frac{b}{bc+b+1}+\frac{c}{ac+c+abc}\); abc = 1 => a;b;c khác 0.
\(\Rightarrow M=\frac{1}{b+1+bc}+\frac{b}{bc+b+1}+\frac{1}{a+1+ab}=\frac{b+1}{bc+b+a}+\frac{abc}{a+abc+ab}\)
\(\Rightarrow M=\frac{b+1}{bc+b+a}+\frac{bc}{1+bc+b}=\frac{bc+b+1}{bc+b+1}=1\)