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

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

Theo bài ra ta có: a.b.c = 1

    =>  a=1;b=1;c=1

Ta có: A = \(\frac{1}{a.b+a+1}\)\(+\frac{1}{b.c+b+1}+\frac{1}{c.a+c+1}\)\(=\frac{1}{1.1+1+1}+\frac{1}{1.1+1+1}\)\(+\frac{1}{1.1+1+1}\)

             \(=\frac{1}{1+1+1}+\frac{1}{1+1+1}+\frac{1}{1+1+1}\)\(=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=\frac{3}{3}=1\)

Vậy A = 1

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

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

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

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

    \(=\frac{c+ac+1}{1+ac+c}=1\)

Ta có:

\(\frac{1}{1+a+a.b}+\frac{1}{1+b+b.c}+\frac{1}{1+c+a.c}\)

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

\(=\frac{1}{1+a+a.b}+\frac{a}{a+a.b+a}+\frac{a.b}{a.b+1+a}\)

\(=\frac{1+a+a.b}{1+a+a.b}=1\)

Đề sai toàn tập, dấu "=" rồi còn tính gì nữa ????