Ta có:

$\dfrac{1}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{1}{abc+bc+b}$

$=\dfrac{abc}{ab+a+abc}+\dfrac{b}{bc+b+1}+\dfrac{1}{1+bc+b}$ (do $abc=1$)

$=\dfrac{abc}{a(bc+b+1)}+\dfrac{b}{bc+b+1}+\dfrac{1}{1+bc+b}$

$=\dfrac{bc}{bc+b+1}+\dfrac{b}{bc+b+1}+\dfrac{1}{1+bc+b}$

$=\dfrac{bc+b+1}{bc+b+1}=1$

(đpcm)

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 

Vì abc=1 nên:

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

Chúc bạn học tốt.

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

\(A=\frac{c}{abc+ac+c}+\frac{ac}{abc\cdot c+abc+ac}+\frac{1}{ac+c+1}\)

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

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

\(A=1\)

ừ

Vậy để mình giúp  

Phải là \(\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ac+bc+1}=1\) thì mới làm đc bạn à 

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

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

\(=\frac{bc}{b+1+bc}+\frac{b}{bc+b+1}+\frac{1}{1+bc+b}\)

\(=\frac{bc+b+1}{bc+b+1}=1\left(đpcm\right)\)