\(B=\frac{1}{1+a+ab}+\frac{a}{a+ab+abc}+\frac{abc}{abc+c+ca}\)

\(=\frac{1}{1+a+ab}+\frac{a}{a+ab+1}+\frac{abc}{c\left(ab+1+a\right)}\)

\(=\frac{1}{1+a+ab}+\frac{a}{a+ab+1}+\frac{ab}{ab+1+a}\)

\(=\frac{1+a+ab}{1+a+ab}=1\)

Có: \(\frac{a}{1+ab}=\frac{b}{1+bc}=\frac{c}{1+ac}\)

Vì a, b, c đôi một khác nhau nên suy ra a, b, c khác 0.

=> \(\frac{1+ab}{a}=\frac{1+bc}{b}=\frac{1+ac}{c}\)

=> \(\frac{1}{a}+b=\frac{1}{b}+c=\frac{1}{c}+a\)

=> \(\hept{\begin{cases}\frac{1}{a}+b=\frac{1}{b}+c\\\frac{1}{b}+c=\frac{1}{c}+a\\\frac{1}{c}+a=\frac{1}{a}+b\end{cases}}\)=> \(\hept{\begin{cases}\frac{b-a}{ab}=c-b\\\frac{c-b}{bc}=a-c\\\frac{a-c}{ac}=b-a\end{cases}}\)

Nhân vế theo vế ta có: \(\frac{\left(b-a\right)\left(c-b\right)\left(a-c\right)}{ab.bc.ac}=\left(c-b\right)\left(a-c\right)\left(b-a\right)\)

=> \(\frac{1}{a^2b^2c^2}=1\)

=> \(\left(abc\right)^2=1\)

=> \(M=abc=\pm1\)

Ta có \(\left\{{}\begin{matrix}B=\text{​​}\frac{1}{1+a+ab}+\frac{1}{1+b+bc}+\frac{1}{1+c+ca}\\abc=1\end{matrix}\right.\)

\(\Rightarrow B=\frac{1}{1+a+ab}+\frac{a}{a+ab+abc}+\frac{ab}{ab+abc+abca}\)

\(\Rightarrow B=\frac{1}{1+a+ab}+\frac{a}{a+ab+1}+\frac{ab}{ab+1+a}\)

\(\Rightarrow B=\frac{1+a+ab}{1+a+ab}=1\)

Vậy B = 1

@@ Học tốt

Chiyuki Fujito

Tái bút : mà bài này còn lận 5 cách nx cơ

\(\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1}=\frac{1}{ab+a+1}+\frac{a}{abc+ab+a}+\frac{ab}{ab.ac+abc+ab}\)

\(=\frac{1}{ab+a+1}+\frac{a}{1+ab+a}+\frac{ab}{a+1+ab}=1\)

\(S=\frac{1}{1+a+ab}+\frac{1}{1+b+bc}+\frac{1}{1+c+ca}=\frac{c}{c+ac+abc}+\frac{ac}{ac+abc+abc^2}+\frac{1}{1+c+ac}=\frac{c}{c+ac+1}+\frac{ac}{ac+1+c}+\frac{1}{1+c+ac}=1\)

S=\(\frac{1}{1+a+ab}+\frac{1}{1+b+bc}+\frac{1}{1+c+ca}\)

=\(\frac{c}{c\left(1+a+ab\right)}+\frac{ac}{ac\left(1+b+bc\right)}+\frac{1}{1+c+ca}\)

=\(\frac{c}{c+ac+abc}+\frac{ac}{ac+abc+abc^2}+\frac{1}{1+c+ca}\)

thay a.b.c=1 ta được 

\(S=\frac{c}{c+ac+1}+\frac{ac}{ac+1+c}+\frac{1}{1+c+a}\)(cộng 3 phân số cùng mẫu c+ac+1)

=\(\frac{c+ac+1}{c+ac+1}=1\)