\(\left\{{}\begin{matrix}ab=c\left(1\right)\\bc=4a\left(2\right)\\ac=9b\left(3\right)\end{matrix}\right.\)
Nhân theo vế của \((1);(2);(3)\) ta có:
\(ab\cdot bc\cdot ac=c\cdot4a\cdot9b\)
\(\Rightarrow a^2b^2c^2=36abc\Rightarrow\left(abc\right)^2=36abc\)
\(\Rightarrow\left(abc\right)^2-36abc=0\)
\(\Rightarrow abc\left(abc-36\right)=0\)\(\Rightarrow\left[{}\begin{matrix}abc=0\\abc-36=0\end{matrix}\right.\)
Do \(a,b,c>0\) nên \(abc=0\) (loại)
Nên ta xét \(abc-36=0\Leftrightarrow abc=36\)
Khi đó \(\left(1\right)\Leftrightarrow abc=c^2\Leftrightarrow36=c^2\Leftrightarrow c=6\left(a,b,c>0\right)\)
\(\left(2\right)\Leftrightarrow abc=4a^2\Leftrightarrow36=4a^2\Leftrightarrow9=a^2\Leftrightarrow a=3\left(a,b,c>0\right)\)
\(\left(3\right)\Leftrightarrow abc=9b^2\Leftrightarrow36=9b^2\Leftrightarrow4=b^2\Leftrightarrow b=2\left(a,b,c>0\right)\)
Vậy \(\left(a;b;c\right)=\left(3;2;6\right)\)
