Có abc=1 nên 
1/(1+a+ab)=abc/(abc+a+ab) 
=abc/[a(1+b+bc)] 
=bc/(1+b+bc) 

1/(1+c+ac)=abc/(abc+c.abc+ac) 
=abc/[ca(1+b+bc)]=b/(1+b+bc) 

=>1/(1+a+ab) + 1/(1+b+bc)+ 1/(1+c+ac) 
=bc/(1+b+bc)+1/(1+b+bc)+b/(1+b+bc) 
=(1+b+bc)/(1+b+bc) 
=1 
=>1/(1+a+ab) + 1/(1+b+bc)+ 1/(1+c+ac)=1

ràu xong

thanks bạn nhiều 

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

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

Ta có \(\frac{1}{ab+a+1}+\frac{b}{bc+b+1}+\frac{1}{abc+bc+b}\)mình chỉnh sửa đề 1 chút , chắc bạn viết sai

\(=\frac{1}{ab+a+1}+\frac{b}{bc+b+1}+\frac{1}{1+bc+b}\)(vì abc=1)

\(=\frac{1}{ab+a+1}+\frac{a.b}{a.\left(bc+b+1\right)}+\frac{a}{a.\left(1+bc+b\right)}\)

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

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

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

\(=1\)

1.\(VT=\frac{c}{abc+ac+c}+\frac{b}{bc+b+abc}+\frac{abc}{abc+bc+b}=\frac{c}{ac+c+1}+\frac{1}{ac+c+1}+\frac{ac}{ac+c+1}=\frac{ac+c+1}{ac+c+1}=1=VP\)

ac+c+1\ac+c+1 =1 

\(\frac{1}{ab+a+1}+\frac{b}{bc+b+1}+\frac{1}{abc+bc+b}\)

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

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

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: \(\frac{\left(b-a\right)\left(c-b\right)\left(a-c\right)}{ab.bc.ac}=\left(c-b\right)\left(a-c\right)\left(b-a\right)\)

=> \(\frac{1}{a^2b^2c^2}=1\)

=> \(\left(abc\right)^2=1\)

=> \(M=abc=\pm1\)