Question

[Mai Chi Lê Vũ]

Cho 3 số a,b,c thỏa mãn a.b.c=1

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

Answer 1

[Nguyễn Minh Tuấn]

*Từ abc=1 => a;b;c khác 0

Khi đó : \(\frac{1}{ab+a+1}\) = \(\frac{1}{ab+a+1}\) .\(\frac{bc}{bc}\) = \(\frac{bc}{ab.bc+abc+bc}\) = \(\frac{bc}{abc.b+abc+bc}\) = \(\frac{bc}{bc+b+1}\)

(do abc=1)

*Do abc = 1 => \(\frac{1}{abc+bc+b}\) = \(\frac{1}{bc+b+1}\)

Khi đó : \(\frac{1}{ab+a+1}\) + \(\frac{b}{bc+b+1}\) + \(\frac{1}{abc+bc+b}\)

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

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

Hay \(\frac{1}{ab+a+1}\) + \(\frac{b}{bc+b+1}\) + \(\frac{1}{abc+bc+b}\) = 1 (đpcm).

*Chú ý : Đây là phương pháp thế số bởi chữ !