Question

[Thuc Anh]

Cho a,b,c thỏa abc=1

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

Answer 1

[Trần Thị Kim Ngân]

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

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

Answer 2

[Trần Việt Anh]

\(S=1\)