Ta có: a2 - b = b2 - c
=> a2 - b2 = -c + b = b - c
=> (a - b)(a + b) = b - c
\(\Rightarrow a+b=\frac{b-c}{a-b}\)
\(\Rightarrow a+b+1=\frac{b-c}{a-b}+1\)
\(\Rightarrow a+b+1=\frac{b-c}{a-b}+\frac{a-b}{a-b}=\frac{b-c+a-b}{a-b}\)
\(\Rightarrow a+b+1=\frac{-c+a}{a-b}=\frac{a-c}{a-b}\)
Ta có: b2 - c = c2 - a
=> b2 - c2 = -a + c = c - a
=> (b - c)(b + c) = c - a
\(\Rightarrow b+c=\frac{c-a}{b-c}\)
\(\Rightarrow b+c+1=\frac{c-a}{b-c}+1=\frac{c-a}{b-c}+\frac{b-c}{b-c}\)
\(\Rightarrow b+c+1=\frac{c-a+b-c}{b-c}=\frac{-a+b}{b-c}\)
\(\Rightarrow b+c+1=\frac{b-a}{b-c}\)
Ta có: a2 - b = c2 - a
=> a2 - c2 = -a + b = b - a
=> (a - c)(a + c) = b - a
\(\Rightarrow a+c=\frac{b-a}{a-c}\)
\(\Rightarrow a+c+1=\frac{b-a}{a-c}+1=\frac{b-a}{a-c}+\frac{a-c}{a-c}\)
\(\Rightarrow c+a+1=\frac{b-a+a-c}{a-c}=\frac{b-c}{a-c}\)
Suy ra: (a + b + 1)(b + c + 1)(c + a + 1)
\(=\frac{a-c}{a-b}.\frac{b-a}{b-c}.\frac{b-c}{a-c}\)
\(=\frac{\left(a-c\right).\left(b-a\right).\left(b-c\right)}{\left(a-b\right).\left(b-c\right).\left(a-c\right)}\)
\(=\frac{b-a}{a-b}\)
= -1
P/s: Ko chắc ạ!
