bài này tôi có thể làm đc nhưng có điều bạn phải tick cho tối đa
`Answer:`
Thêm điều kiện `a,b,c\ne0` nhé.
\(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\)
\(\Rightarrow\frac{a+b}{ab}=\frac{b+c}{bc}=\frac{c+a}{ca}\)
\(\Rightarrow\frac{a}{ab}+\frac{b}{ab}=\frac{b}{bc}+\frac{c}{bc}=\frac{c}{ca}+\frac{a}{ca}\)
\(\Rightarrow\frac{1}{b}+\frac{1}{a}=\frac{1}{c}+\frac{1}{b}=\frac{1}{a}+\frac{1}{c}\)
\(\Rightarrow\hept{\begin{cases}\frac{1}{b}+\frac{1}{a}=\frac{1}{c}+\frac{1}{b}\\\frac{1}{c}+\frac{1}{b}=\frac{1}{a}+\frac{1}{c}\\\frac{1}{a}+\frac{1}{c}=\frac{1}{b}+\frac{1}{a}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}\frac{1}{a}=\frac{1}{c}\\\frac{1}{b}=\frac{1}{a}\\\frac{1}{c}=\frac{1}{b}\end{cases}}\)
\(\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\)
`=>a=b=c`
Lúc này `M=\frac{ab+bc+ca}{a^2+b^2+c^2}=\frac{1.1+1.1+1.1}{1^2+1^2+1^2}=3/3=1`
Ta có: `ab/(a+b) = bc/(b+c) = ca/(c+a)`
`=> (a+b)/ab = (b+c)/bc = (c+a)/ca`
`=> a/ab + b/ab = b/bc + c/bc = c/ca + a/ca`
`=> 1/b + 1/a = 1/c + 1/b = 1/a + 1/c`
=> \(\left\{{}\begin{matrix}\dfrac{1}{b}+\dfrac{1}{a}=\dfrac{1}{c}+\dfrac{1}{b}\\\dfrac{1}{c}+\dfrac{1}{b}=\dfrac{1}{a}+\dfrac{1}{c}\\\dfrac{1}{a}+\dfrac{1}{c}=\dfrac{1}{b}+\dfrac{1}{a}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{1}{a}=\dfrac{1}{b}\\\dfrac{1}{b}=\dfrac{1}{a}\\\dfrac{1}{c}=\dfrac{1}{b}\end{matrix}\right.\)
`=>1/a = 1/b = 1/c`
`=> a = b = c`
Đặt:`a = b = c = 1`
Ta có: M=\(\dfrac{ab+bc+ca}{a^2+b^2+c^2}=\dfrac{1.1+1.1+1.1}{1^2+1^2+1^2}=\dfrac{3}{3}=1\)
Vậy M=1
HT
