Từ M=\(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\)
\(\Rightarrow\frac{a+b}{ab}=\frac{b+c}{bc}=\frac{c+a}{ca}\)
\(\Rightarrow\frac{a}{ab}+\frac{b}{ab}=\frac{b}{bc}+\frac{c}{bc}=\frac{c}{ca}+\frac{a}{ca}\)
\(\Rightarrow\frac{1}{b}+\frac{1}{a}=\frac{1}{c}+\frac{1}{b}=\frac{1}{a}+\frac{1}{c}\)
\(\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\)
\(\Rightarrow a=b=c\)
Ta có: \(M=\frac{ab+bc+ca}{a^2+b^2+c^2}=\frac{a^2+a^2+a^2}{a^2+a^2+a^2}=1\)
Vậy M= 1
Ta có \(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{a+c}\Rightarrow\frac{a+b}{ab}=\frac{b+c}{bc}=\frac{a+c}{ac}\)
\(\Rightarrow\frac{a}{ab}+\frac{b}{ab}=\frac{b}{bc}+\frac{c}{bc}=\frac{c}{ca}+\frac{a}{ca}\Leftrightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{b}+\frac{1}{c}=\frac{1}{c}+\frac{1}{a}\)
Có \(\frac{1}{a}+\frac{1}{b}=\frac{1}{b}+\frac{1}{c}\Leftrightarrow\frac{1}{a}=\frac{1}{c}\left(1\right)\) và \(\frac{1}{b}+\frac{1}{c}=\frac{1}{c}+\frac{1}{a}\Leftrightarrow\frac{1}{b}=\frac{1}{c}\left(2\right)\)
Từ \(\left(1\right)\left(2\right)\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\) hay \(a=b=c\)
Vậy \(M=\frac{ab+bc+ca}{a^2+b^2+c^2}=\frac{a^2+b^2+c^2}{a^2+b^2+c^2}=1\)
