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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
\(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ca}{c+a}\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{c}+\dfrac{1}{a}\)
\(\Rightarrow\dfrac{1}{a}=\dfrac{1}{b}=\dfrac{1}{c}=\dfrac{1+1+1}{a+b+c}=\dfrac{3}{a+b+c}=\dfrac{3}{1}=3\)
\(\Rightarrow a=b=c=\dfrac{1}{3}\)
\(\Rightarrow A=\dfrac{a^3\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2}=a^3=\left(\dfrac{1}{3}\right)^3=\dfrac{1}{27}\)
\(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\)
\(\Rightarrow\frac{a+b}{ab}=\frac{b+c}{bc}=\frac{c+a}{ca}\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{b}+\frac{1}{c}=\frac{1}{c}+\frac{1}{a}\)
\(\frac{\Rightarrow1}{a}=\frac{1}{b}=\frac{1}{c}\Rightarrow a=b=c\)
Thay vào M ta có
\(\frac{a^2+a^2+a^2}{a^2+a^2+a^2}=1\)
P/s : hỏi từng câu thôi
\(a,\dfrac{3}{a+b}=\dfrac{2}{b+c}=\dfrac{1}{c+a}\\ \Rightarrow\dfrac{a+b}{3}=\dfrac{b+c}{2}=\dfrac{c+a}{1}=\dfrac{2\left(a+b+c\right)}{6}=\dfrac{a+b+c}{3}\\ \Rightarrow\dfrac{a+b}{3}=\dfrac{a+b+c}{3}\\ \Rightarrow3\left(a+b+c\right)=3\left(a+b\right)\\ \Rightarrow3\left(a+b\right)+3c=3\left(a+b\right)\\ \Rightarrow3c=0\\ \Rightarrow c=0\)
Vậy \(P=\dfrac{a+b-2019c}{a+b+2018c}=\dfrac{a+b}{a+b}=1\)
