Question

Cho a*b*c=1

chứng minh rằng \(\frac{2a}{ab+a+1}+\frac{2b}{bc+b+1}+\frac{2c}{ac+c+1}=2\)

Answer 1

bài này easy thôi:

chia cả 2 vế cho 2 ta được:

\(P=\frac{a}{ab+a+1}+\frac{b}{bc+c+1}+\frac{c}{ac+c+1}=1\)

Thật vậy:ta có:\(abc=1\Leftrightarrow a=\frac{1}{bc}\)

\(\Rightarrow P=\frac{\frac{1}{bc}}{\frac{1}{bc}b+\frac{1}{bc}+1}+\frac{b}{bc+b+1}+\frac{c}{\frac{1}{bc}+c+1}\)

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

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

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

\(\Rightarrowđpcm\)