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Ta có :
\(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}=\frac{ab-bc}{\left(a+b\right)-\left(b+c\right)}=\frac{bc-ca}{\left(b+c\right)-\left(c+a\right)}=\frac{ab-ca}{\left(a+b\right)-\left(c+a\right)}\)
\(\Rightarrow a=b=c\)
\(\Rightarrow Q=\frac{ab^2+bc^2+ca^2}{a^3+b^3+c^3}=1\)
\(\frac{ab+ac}{2}=\frac{bc+ab}{3}=\frac{ca+bc}{4}\)
( ta lần lược lấy - (1) + (2) + (3) = (1) - (2) + (3) = (1) + (2) - (3) được)
\(=\frac{2bc}{5}=\frac{2ca}{3}=\frac{2ab}{1}\)
Ta thấy rằng a,b,c không thể = 0 vì như vậy thì a + b + c \(\ne69\)
\(\Rightarrow\hept{\begin{cases}a=\frac{c}{5}\\b=\frac{c}{3}\end{cases}}\)
Thế vào: a + b + c = 69
\(\Leftrightarrow\frac{c}{5}+\frac{c}{3}+c=69\)
\(\Rightarrow c=45\)
\(\Rightarrow\hept{\begin{cases}a=9\\b=15\end{cases}}\)
Từ \(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\) => \(\frac{a+b}{ab}=\frac{b+c}{bc}=\frac{c+a}{ca}\) => \(\frac{a}{ab}+\frac{b}{ab}=\frac{b}{bc}+\frac{c}{bc}=\frac{c}{ca}+\frac{a}{ca}\)
=> \(\frac{1}{b}+\frac{1}{a}=\frac{1}{c}+\frac{1}{b}=\frac{1}{a}+\frac{1}{c}\) => \(\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\) => a = b = c
Vậy B = \(\frac{a.a^2+b.b^2+c.c^2}{a^3+b^3+c^3}=\frac{a^3+b^3+c^3}{a^3+b^3+c^3}=1\)
\(\frac{1}{1+a+ab}+\frac{1}{1+b+bc}+\frac{1}{1+c+ac}\)
\(=\frac{abc}{abc+a\times abc+ab}+\frac{abc}{abc+b+bc}+\frac{1}{1+c+ac}\)
\(=\frac{abc}{ab\left(c+ac+1\right)}+\frac{abc}{b\left(ac+1+c\right)}+\frac{1}{1+c+ac}\)
\(=\frac{c}{c+ac+1}+\frac{ac}{ac+1+c}+\frac{1}{1+c+ac}\)
\(=\frac{c+ac+1}{c+ac+1}\)
= 1
\(\frac{ab}{a+b}=\frac{ac}{a+c}=\frac{bc}{b+c}\Rightarrow\frac{abc}{c\left(a+b\right)}=\frac{abc}{b\left(a+c\right)}=\frac{abc}{a\left(b+c\right)}\)
\(\Rightarrow c\left(a+b\right)=b\left(a+c\right)\Leftrightarrow ac+bc=ab+bc\Rightarrow ac=ab\Rightarrow c=b\) (1)
\(\Rightarrow b\left(a+c\right)=a\left(b+c\right)\Leftrightarrow ab+bc=ab+ac\Rightarrow bc=ac\Rightarrow b=a\) (2)
\(\Rightarrow c\left(a+b\right)=a\left(b+c\right)\Leftrightarrow ac+bc=ab+ac\Rightarrow bc=ab\Rightarrow c=a\) (3)
Từ (1) ; (2) ; (3) => \(a=b=c\) (ĐPCM)