Question

Cho abc=1

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

Answer 1

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