Ta có: \(\dfrac{3}{a}=\dfrac{4}{b}=\dfrac{9}{c}\)
nên \(\dfrac{3}{a}=\dfrac{4}{b}=\dfrac{9}{c}=\dfrac{3+4+9}{a+b+c}=\dfrac{16}{a+b+c}\)
Ta có: \(a+b+c=\dfrac{3}{a}=\dfrac{4}{b}=\dfrac{9}{c}\)
\(\Leftrightarrow a+b+c=\dfrac{16}{a+b+c}\)
\(\Leftrightarrow\left(a+b+c\right)^2=16\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=4\\a+b+c=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{a}=\dfrac{4}{b}=\dfrac{9}{c}=4\\\dfrac{3}{a}=\dfrac{4}{b}=\dfrac{9}{c}=-4\end{matrix}\right.\)
Trường hợp 1: \(\dfrac{3}{a}=\dfrac{4}{b}=\dfrac{9}{c}=4\)
nên \(\left\{{}\begin{matrix}a=\dfrac{3}{4}\\b=1\\c=\dfrac{9}{4}\end{matrix}\right.\)
Trường hợp 2: \(\dfrac{3}{a}=\dfrac{4}{b}=\dfrac{9}{c}=-4\)
nên \(\left\{{}\begin{matrix}a=\dfrac{-3}{4}\\b=-1\\c=\dfrac{-9}{4}\end{matrix}\right.\)
Vậy: \(\left(a,b,c\right)\in\left\{\left(\dfrac{3}{4};1;\dfrac{9}{4}\right);\left(-\dfrac{3}{4};-1;-\dfrac{9}{4}\right)\right\}\)
