Ta có: a . b = 2 ; b . c = 3 ; c . a = 54
\(\Rightarrow\) a . b . b . c . c . a = 2 . 3 . 54 = 324
hay a2 . b2 . c2 = 324 hay ( a . b . c )2 = 324
\(\Rightarrow\) a . b . c = 18
hay 2 . c = 18 \(\Rightarrow\) c = 9
hay a . 3 = 18 \(\Rightarrow\) a = 6
hay b . 54 = 18 \(\Rightarrow\) b = 1/3
Giải:
Có: \(ab=2\); \(bc=3\); \(ca=54\)
\(\Leftrightarrow ab.bc.ca=2.3.54\)
\(\Leftrightarrow\left(abc\right)^2=324\)
\(\Leftrightarrow abc=\sqrt{324}=18\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{abc}{bc}\\b=\dfrac{abc}{ca}\\c=\dfrac{abc}{ab}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{18}{3}\\b=\dfrac{18}{54}\\c=\dfrac{18}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=6\\b=\dfrac{1}{3}\\c=9\end{matrix}\right.\)
Vậy \(a=6\); \(b=\dfrac{1}{3}\) và \(c=9\).
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