Lời giải:
Dựa vào điều kiện $abc=1$ ta có:

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

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

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

Ta có đpcm.

Ta có: \(a.b.c=1\)

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

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

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

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

\(=1.\)

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

Chúc bạn học tốt!

Có : 1/ab+a+1 = abc/ab+a+abc = bc/b+1+bc

1/abc+bc+b  = 1/1+bc+b

=> 1/ab+a+1 + b/bc+b+1 + 1/abc+bc+b = bc/bc+c+1 + b/bc+b+1 + 1/bc+b+1 = bc+b+1/bc+b+1 = 1

=> ĐPCM

k mk nha

Có : 1/ab+a+1 = abc/ab+a+abc = bc/b+1+bc

1/abc+bc+b  = 1/1+bc+b

=> 1/ab+a+1 + b/bc+b+1 + 1/abc+bc+b = bc/bc+c+1 + b/bc+b+1 + 1/bc+b+1 = bc+b+1/bc+b+1 = 1

=> ĐPCM