Question

Cho a;b;c là 3 số thỏa mãn: abc = 1. Tính S = \(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ac+c+1}\)

 

Answer 1

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

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

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

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

\(=1\)

Answer 2

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

        c/ac+c+1 = bc/abc+bc+b = bc/bc+b+1

=> S = 1+b+bc/bc+b+1 = 1

Vậy S = 1

Tk mk nha