Giải:
Ta có: \(\left\{{}\begin{matrix}ab=-18\left(1\right)\\bc=15\left(2\right)\\ac=-30\left(3\right)\end{matrix}\right.\)
Nhân \(\left(1\right);\left(2\right);\left(3\right)\) theo vế ta được:
\(abbcac=-18.15.\left(-30\right)=8100\)
\(\Rightarrow a^2b^2c^2=\left(abc\right)^2=8100\)
\(\Leftrightarrow abc=\sqrt{8100}=\pm90\)
\(\Rightarrow\left\{{}\begin{matrix}c=\dfrac{abc}{ab}=\dfrac{\pm90}{-18}=\pm5\\a=\dfrac{abc}{bc}=\dfrac{\pm90}{15}=\pm6\\b=\dfrac{abc}{ac}=\dfrac{\pm90}{-30}=\pm3\end{matrix}\right.\)
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