có \(ab=\frac{3}{5}\Leftrightarrow a=\frac{3}{5b}\)
có \(bc=\frac{4}{5}\Rightarrow c=\frac{4}{5b}\)
mà \(ac=\frac{4}{5}\Leftrightarrow\frac{3}{5b}.\frac{4}{5b}=\frac{3}{4}\Leftrightarrow\frac{12}{25.b^2}=\frac{3}{4}\Leftrightarrow12.4=3.25.b^2\)
\(\Leftrightarrow\frac{48}{75}=b^2\Leftrightarrow b^2=\frac{16}{25}\Leftrightarrow b=\pm\frac{4}{5}\)
với \(b=\frac{4}{5}\)thì \(a=3:\left(5.\frac{4}{5}\right)=3:4=\frac{3}{4}\)
\(c=4:\left(5.\frac{4}{5}\right)=4:4=1\)
với \(b=-\frac{4}{5}\)thì \(a=3:\left(5.\frac{-4}{5}\right)=3:-4=-\frac{3}{4}\)
\(c=4:\left(5.\frac{-4}{5}\right)=4:-4=-1\)
vậy \(\left(a;b;c\right)\in\left\{\left(\frac{4}{5};\frac{3}{4};1\right);\left(\frac{-4}{5};\frac{-3}{4};-1\right)\right\}\)
