Question

Cho a,b,c thỏa mãn:abc=1

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

Answer 1

Ta có \(\frac{1}{ab+a+1}+\frac{b}{bc+b+1}+\frac{1}{abc+bc+b}\)mình chỉnh sửa đề 1 chút , chắc bạn viết sai

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

\(=\frac{1}{ab+a+1}+\frac{a.b}{a.\left(bc+b+1\right)}+\frac{a}{a.\left(1+bc+b\right)}\)

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

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

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

\(=1\)