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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
ab = bc
\(\Rightarrow\) a = c (1)
bc = cd
\(\Rightarrow\) b = d (2)
cd = de
\(\Rightarrow\) c = e (3)
de = ea
\(\Rightarrow\) d = a (4)
ea = ab
\(\Rightarrow\) e = b (5)
Từ (1), (2), (3), (4), (5) \(\Rightarrow\) a = b = c = d = e
\(\Rightarrow\) ĐPCM
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ca}{c+a}\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{c}+\dfrac{1}{a}\)
\(\Rightarrow\dfrac{1}{a}=\dfrac{1}{b}=\dfrac{1}{c}=\dfrac{1+1+1}{a+b+c}=\dfrac{3}{a+b+c}=\dfrac{3}{1}=3\)
\(\Rightarrow a=b=c=\dfrac{1}{3}\)
\(\Rightarrow A=\dfrac{a^3\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2}=a^3=\left(\dfrac{1}{3}\right)^3=\dfrac{1}{27}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài này tui làm rùi mà.
\(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}=\frac{1}{\frac{1}{a}+\frac{1}{b}}=\frac{1}{\frac{1}{b}+\frac{1}{c}}=\frac{1}{\frac{1}{c}+\frac{1}{a}}\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{b}+\frac{1}{c}=\frac{1}{c}+\frac{1}{a}\Leftrightarrow a=b=c\)
\(\Leftrightarrow M=\frac{ab+bc+ca}{a^2+b^2+c^2}=\frac{3a^2}{3a^2}=1\)