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

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)

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

dung roi day no lam theo cach cua to

1/1+a+ab  +1/1+b+bc  +1/1+c+ac

=1/a+1+ab  +a/a+ab+abc  +ab/ab+abc+acab

=1/a+1+ab  +a/a+ab+1  +ab/ab+1+a

=1+a+ab/1+a+ab

=1

vậy 1/a+1+ab  +1/1+b+bc  +1/1+c+ca =1(đpcm)

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

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

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

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

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

thay a.b.c=1 Ta đc:

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

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

tick cho mk vs nhé

Với \(a=b=c=0\Leftrightarrow S=abc=0\)

Với \(a,b,c\ne0\)

Ta có \(\dfrac{a}{1+ab}=\dfrac{b}{1+bc}=\dfrac{c}{1+ac}\Leftrightarrow\dfrac{1+ab}{a}=\dfrac{1+bc}{b}=\dfrac{1+ac}{c}\)

\(\Leftrightarrow\dfrac{1}{a}+b=\dfrac{1}{b}+c=\dfrac{1}{c}+a\)

\(\Leftrightarrow\left\{{}\begin{matrix}a-b=\dfrac{1}{a}-\dfrac{1}{c}=\dfrac{c-a}{ac}\\b-c=\dfrac{1}{b}-\dfrac{1}{a}=\dfrac{a-b}{ab}\\c-a=\dfrac{1}{c}-\dfrac{1}{b}=\dfrac{b-c}{bc}\end{matrix}\right.\)

Nhân vế theo vế ta đc \(\left(a-b\right)\left(b-c\right)\left(c-a\right)=\dfrac{\left(a-b\right)\left(b-c\right)\left(c-a\right)}{ab\cdot bc\cdot ca}\)

\(\Leftrightarrow\left(abc\right)^2=1\Leftrightarrow\left[{}\begin{matrix}abc=1\\abc=-1\end{matrix}\right.\)

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