\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\)
\(\Rightarrow\dfrac{a^2}{4}=\dfrac{b^2}{9}=\dfrac{c^2}{16}=\dfrac{3b^2}{27}=\dfrac{2c^2}{32}=\dfrac{a^2+3b^2-2c^2}{4+27-32}=\dfrac{-16}{-1}=16\)
\(\Rightarrow\left\{{}\begin{matrix}a^2=64\\b^2=144\\c^2=256\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a=\pm8\\b=\pm12\\c=\pm16\end{matrix}\right.\)
Vậy \(\left(a;b;c\right)\in\left\{\left(8;12;16\right),\left(-8;-12;-16\right)\right\}\)
Cách khác:
Đặt \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=2k\\b=3k\\c=4k\end{matrix}\right.\)
Ta có: \(a^2+3b^2-2c^2=-16\)
\(\Leftrightarrow4k^2+27k^2-32k^2=-16\)
\(\Leftrightarrow k^2=16\)
Trường hợp 1: k=4
\(\Leftrightarrow\left\{{}\begin{matrix}a=2k=8\\b=3k=12\\c=4k=16\end{matrix}\right.\)
Trường hợp 2: k=-4
\(\Leftrightarrow\left\{{}\begin{matrix}a=2k=-8\\b=3k=-12\\c=4k=-16\end{matrix}\right.\)
