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

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

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

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

\(\Leftrightarrow a=b=c\)

\(\Rightarrow P=1\)

ta có \(\left\{{}\begin{matrix}\dfrac{ab}{a+b}=\dfrac{ac}{a+c}\\\dfrac{ab}{a+b}=\dfrac{bc}{b+c}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a.\dfrac{b}{a+b}=a.\dfrac{c}{c+a}\\b.\dfrac{a}{a+b}=b.\dfrac{c}{b+c}\end{matrix}\right.\)

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

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

\(\Rightarrow P=\dfrac{ab^2+bc^2+ca^2}{a^3+b^3+c^3}=\dfrac{a^3+a^3+a^3}{a^3+a^3+a^3}=1\)

a=1 ; b=2 ;c=3

Ta có : a.b.a.c = 2.3 = 6 ; a^2 .b.c = 6 mà b.c = 6 suy ra a^2 = 1.

Vậy a = 1. Từ đó suy ra : b = 2 ; c = 3

B

B. A.B+A.C

\(ab=2,bc=3,ac=54\)

\(\Rightarrow ab.bc.ac=2.3.54\)

\(\Rightarrow\left(abc\right)^2=324\)

\(\Rightarrow\left(abc\right)^2=18^2=\left(-18\right)^2\)

+)\(abc=18\)

\(\Rightarrow a=18:3=6\)

\(\Rightarrow b=18:54=\frac{1}{3}\)

\(\Rightarrow c=18:2=9\)

+)\(abc=-18\)

\(\Rightarrow a=-18:3=-6\)

\(\Rightarrow b=-18:54=\frac{-1}{3}\)

\(\Rightarrow c=-18:2=-9\)

Vậy :\(a\in\left(6;-6\right)\)

        \(b\in\left(\frac{1}{3};\frac{-1}{3}\right)\)

         \(c\in\left(9;-9\right)\)

Vì a, b, c là 3 số dương => a > 0 ; b > 0 ; c > 0

Ta có : a.b = c => b.c = b.a.b = a.b² = 4a => b² = 4 => b = 2 (vì b > 0)

b.c = 4a => 2.c = 4a => c = 2a

a.c = 9b => a.2a = 9.2 => 3a = 18 => a = 6

=> a.c = 9b => 6.c = 18 => c = 3 

Vậy a = 6 , b = 2 , c = 3 thì thỏa mãn đề bài