Ta có: ab=2 (1)
bc=6 (2)
ac=3 (3)
Nhân (1);(2);(3) :
a.c.b.c.a.c=\(a^2.b^2.c^2=\left(abc\right)^2\)=2.6.3=36
\(\Rightarrow\) abc=6
Mà ab=2
=>c=3
bc=6
=>a=1
ac=3
=>b=2
Ta có:
\(\left\{{}\begin{matrix}ab=2\\bc=6\\ac=3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}ab=2\\\dfrac{bc}{ac}=\dfrac{6}{3}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}ab=2\\\dfrac{b}{a}=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}ab=2\\b=2a\end{matrix}\right.\)
\(\Leftrightarrow2a^2=2\Leftrightarrow a^2=1\)
Mà:
\(b=2a\Leftrightarrow b^2=4a^2=4\)
\(ac=3\Leftrightarrow a^2c^2=9\Leftrightarrow c^2=9\)
\(\Leftrightarrow Q=a^2+b^2+c^2=1+4+9=14\)
Vậy \(Q=14\)
