Question

Cho a,b,c thỏa mãn abc=1

Tính B= \(\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}\)

Answer 1

B=\(\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}\)

=\(\dfrac{a}{ab+a+abc}+\dfrac{b}{bc+b+abc}+\dfrac{c}{ac+c+1}\)

= \(\dfrac{a}{a\left(b+1+bc\right)}\)\(+\dfrac{b}{b \left(c+1+ac\right)}+\dfrac{c}{c\left(ac+1+c\right)}\)

=\(\dfrac{1}{b+1+bc}+\dfrac{1}{ac+c+1}+\dfrac{1}{ac+c+1}\)

=\(\dfrac{ac}{ac+c+1}+\dfrac{1}{ac+c+1}+\dfrac{c}{ac+c+1}\)

=\(\dfrac{ac+1+c}{ac+1+c}=1\)

Answer 2

\(B=\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}=\dfrac{ac}{abc+ac+c}+\dfrac{abc}{abcc+abc+ac}+\dfrac{c}{ac+c+1}.Thay:abc=1,ta,được:B=\dfrac{ac}{ac+c+1}+\dfrac{1}{ac+c+1}+\dfrac{c}{ac+c+1}\Rightarrow B=\dfrac{ac+c+1}{ac+c+1}=1.Vậy:B=1\)