\(Cho\)\(\frac{a+b}{ab}=\frac{b+c}{bc}=\frac{c+a}{ca}\)\(T\text{ính}\)\(gi\text{á}\)\(tr\text{ị}\)\(c\text{ủ}a\)\(M=\frac{ab+bc+ca}{a^2+b^2+c^2}\)
\(gi\text{úp}\)\(m\text{ình}\)\(v\text{ới}\)\(mai\)\(m\text{ình}\)\(\text{đ}i\)\(h\text{ọc}\)\(r\text{ồi}\)
\(c\text{ảm}\)\(\text{ơ}n\)\(nhi\text{ều}\)
Ta có :\(\frac{a+b}{ab}=\frac{b+c}{bc}=\frac{c+a}{ca}\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{b}+\frac{1}{c}=\frac{1}{c}+\frac{1}{a}\)
\(\Rightarrow\frac{1}{a}=\frac{1}{c};\frac{1}{b}=\frac{1}{a};\frac{1}{c}=\frac{1}{b}\)
\(\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\)
\(\Rightarrow a=b=c\)
\(\Rightarrow ab=bc=ca=a^2=b^2=c^2\)
\(\Rightarrow ab+bc+ca=a^2+b^2+c^2\)
\(\Rightarrow\frac{ab+bc+ca}{a^2+b^2+c^2}=1\)
Vậy M=1