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\(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ca}{c+a}\)
=> \(\dfrac{abc}{ac+bc}=\dfrac{abc}{ab+ac}=\dfrac{abc}{bc+ab}\)
=> ac + bc = ab + ac = bc + ab (do abc \(\ne0\))
=> ac + bc - ab - ac = 0
=> bc - ab = 0
=> b(c - a) = 0
Mà b \(\ne0\) nên c - a = 0 => c = a
Tương tự ta có: a = b
Từ đó có: a = b = c
Thay vào M được:
\(M=\dfrac{a^2+a^2+a^2}{a^2+a^2+a^2}=1\)
Do \(a,b,c\ne0\)
\(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ac}{a+c}\Rightarrow\dfrac{a+b}{ab}=\dfrac{b+c}{bc}=\dfrac{a+c}{ac}\)
\(\Rightarrow\dfrac{a}{ab}+\dfrac{b}{ab}=\dfrac{b}{bc}+\dfrac{c}{bc}=\dfrac{a}{ac}+\dfrac{c}{ac}\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a}+\dfrac{1}{c}\) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{b}+\dfrac{1}{c}\\\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a}+\dfrac{1}{c}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a}=\dfrac{1}{c}\\\dfrac{1}{b}=\dfrac{1}{a}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=c\\b=a\end{matrix}\right.\) \(\Rightarrow a=b=c\)
\(\Rightarrow M=\dfrac{a.a+a.a+a.a}{a^2+a^2+a^2}=\dfrac{3a^2}{3a^2}=1\)
3a-b=1/2(a+b)
=>6a-2b=a+b
=>5a=3b
=>a/3=b/5=k
=>a=3k; b=5k
\(A=\dfrac{a^{2022}+3^{2022}}{b^{2022}+5^{2022}}\)
\(=\dfrac{3^{2022}\left(k^{2022}+1\right)}{5^{2022}\left(k^{2022}+1\right)}=\left(\dfrac{3}{5}\right)^{2022}\)