Ta có:\(\dfrac{ab}{bc}=\dfrac{2}{6}\Leftrightarrow\dfrac{a}{c}=\dfrac{1}{3}\Leftrightarrow c=3a\)
ac=3 mà c=3a
=>3a2=3
<=>a2=1
=>a=1 hoặc a=-1
*)a=1
ab=2=>b=2
ac=3=>c=3
*)a=-1
ab=2=>b=-2
ac=3=>c=-3
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Ta có: \(\left\{{}\begin{matrix}ab=2\left(1\right)\\bc=6\left(2\right)\\ac=3\left(3\right)\end{matrix}\right.\)
Nhân các vế \(\left(1\right);\left(2\right);\left(3\right)\) với nhau ta được:
\(abbcac=2.6.3\Leftrightarrow a^2b^2c^2=36\)
\(\Leftrightarrow\left(abc\right)^2=36\Rightarrow abc=\sqrt{36}=\pm6\)
Ta có 2 trường hợp:
Trường hợp 1: Nếu \(abc=6\)
\(\Rightarrow\left\{{}\begin{matrix}ab=2\\bc=6\\ac=3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}c=\dfrac{abc}{ab}=\dfrac{6}{2}=3\\a=\dfrac{abc}{bc}=\dfrac{6}{6}=1\\b=\dfrac{abc}{ac}=\dfrac{6}{3}=2\end{matrix}\right.\)
Trường hợp 2: Nếu \(abc=-6\)
\(\Rightarrow\left\{{}\begin{matrix}ab=2\\bc=6\\ac=3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}c=\dfrac{abc}{ab}=\dfrac{-6}{2}=-3\\a=\dfrac{abc}{bc}=\dfrac{-6}{6}=-1\\b=\dfrac{abc}{ac}=\dfrac{-6}{3}=-2\end{matrix}\right.\)
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