Question

cho a*b*c=1, tính:

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

Answer 1

Từ a.b.c=1 ta có:

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