Ta có : \(3a=-2b\Leftrightarrow a=\dfrac{-2b}{3}\)
\(5c=-2b\Leftrightarrow c=\dfrac{-2b}{5}\)
Thay vào \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=2\), ta có :
\(\Leftrightarrow\dfrac{1}{-\dfrac{2b}{3}}+\dfrac{1}{b}+\dfrac{1}{-\dfrac{2b}{5}}=2\)
\(\Leftrightarrow\dfrac{-3}{2b}+\dfrac{1}{b}+\dfrac{-5}{2b}=2\)
\(\Leftrightarrow\dfrac{-3+2-5}{2b}=2\)
\(\Leftrightarrow-3b=2\)
\(\Leftrightarrow b=-\dfrac{2}{3}\)
Ta có : \(a=\dfrac{-2b}{3}=\dfrac{-2.\left(-\dfrac{2}{3}\right)}{3}=\dfrac{4}{9}\)
\(c=\dfrac{-2b}{5}=\dfrac{-2.\left(-\dfrac{2}{3}\right)}{5}=\dfrac{4}{15}\)
Vậy \(a=\dfrac{4}{9},b=-\dfrac{2}{3},c=\dfrac{4}{15}\).
