Từ \(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ca}{c+a}\)

\(\Rightarrow\dfrac{a+b}{ab}=\dfrac{b+c}{bc}=\dfrac{c+a}{ca}\)

\(\Rightarrow\dfrac{a}{ab}+\dfrac{b}{ab}=\dfrac{b}{bc}+\dfrac{c}{bc}=\dfrac{c}{ca}+\dfrac{a}{ca}\)

\(\Rightarrow\dfrac{1}{b}+\dfrac{1}{a}=\dfrac{1}{c}+\dfrac{1}{b}=\dfrac{1}{a}+\dfrac{1}{c}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{b}+\dfrac{1}{a}=\dfrac{1}{c}+\dfrac{1}{b}\\\dfrac{1}{c}+\dfrac{1}{b}=\dfrac{1}{a}+\dfrac{1}{c}\\\dfrac{1}{a}+\dfrac{1}{c}=\dfrac{1}{b}+\dfrac{1}{a}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a}=\dfrac{1}{c}\\\dfrac{1}{b}=\dfrac{1}{a}\\\dfrac{1}{c}=\dfrac{1}{b}\end{matrix}\right.\)

\(\Rightarrow\dfrac{1}{a}=\dfrac{1}{b}=\dfrac{1}{c}\Rightarrow a=b=c\)

Khi đó: \(M=\dfrac{ab+bc+ca}{a^2+b^2+c^2}=\dfrac{1\cdot1+1\cdot1+1\cdot1}{1^2+1^2+1^2}=\dfrac{3}{3}=1\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{ab+ac}{2}=\frac{ba+bc}{3}=\frac{ca+cb}{4}=\frac{\left(ab+ac\right)+\left(ba+bc\right)-\left(ca+cb\right)}{2+3-4}=\frac{2ab}{1}\)

Tương tự \(\frac{ab+ac}{2}=\frac{bc+ba}{3}=\frac{ca+cb}{4}=\frac{2bc}{5}\)

\(\frac{ab+ac}{2}=\frac{ba+bc}{3}=\frac{ca+cb}{4}=\frac{2ac}{3}\)

Do đó \(\frac{2ab}{1}=\frac{2bc}{5}\Rightarrow\frac{a}{1}=\frac{c}{5}\Rightarrow\frac{a}{3}=\frac{c}{15}\)

\(\frac{2bc}{5}=\frac{2ac}{3}\Rightarrow\frac{b}{5}=\frac{a}{3}\)

Do vậy \(\frac{a}{3}=\frac{b}{5}=\frac{c}{15}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

$\frac{ab+ac}{2}=\frac{ba+bc}{3}=\frac{ca+cb}{4}=\frac{\left(ab+ac\right)+\left(ba+bc\right)-\left(ca+cb\right)}{2+3-4}=\frac{2ab}{1}$

Tương tự $\frac{ab+ac}{2}=\frac{bc+ba}{3}=\frac{ca+cb}{4}=\frac{2bc}{5}$

$\frac{ab+ac}{2}=\frac{ba+bc}{3}=\frac{ca+cb}{4}=\frac{2ac}{3}$

Do đó $\frac{2ab}{1}=\frac{2bc}{5}\Rightarrow \frac{a}{1}=\frac{c}{5}\Rightarrow \frac{a}{3}=\frac{c}{15}$

$\frac{2bc}{5}=\frac{2ac}{3}\Rightarrow \frac{b}{5}=\frac{a}{3}$

Do vậy $\frac{a}{3}=\frac{b}{5}=\frac{c}{15}$

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