Question

Cho các số a, b, c thỏa mã a.b.c = 1

Tính A = \(\frac{1}{a.b+a+1}+\frac{1}{b.c+b+1}+\frac{1}{c.a+c+1}\)

Answer 1

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

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

Answer 2

Theo bài ra ta có: a.b.c = 1

    =>  a=1;b=1;c=1

Ta có: A = \(\frac{1}{a.b+a+1}\)\(+\frac{1}{b.c+b+1}+\frac{1}{c.a+c+1}\)\(=\frac{1}{1.1+1+1}+\frac{1}{1.1+1+1}\)\(+\frac{1}{1.1+1+1}\)

             \(=\frac{1}{1+1+1}+\frac{1}{1+1+1}+\frac{1}{1+1+1}\)\(=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=\frac{3}{3}=1\)

Vậy A = 1