Đặt \(\frac{a}{3}=\frac{b}{4}=\frac{c}{5}=k\)
\(=>\hept{\begin{cases}a=3k\\b=4k\\c=5k\end{cases}}\)
Khi đó : \(2a^2+2b^2-3c^2=-100\)
\(< =>2\left(3k\right)^2+2\left(4k\right)^2-3\left(5k\right)^2=-100\)
\(< =>2.9.k^2+2.16.k^2-3.25.k^2=-100\)
\(< =>19k^2+32k^2-75k^2=-100\)
\(< =>k^2\left(51-75\right)=-100\)
\(< =>-24k^2=-100\)
\(< =>k^2=\frac{25}{6}\)\(< =>k=\pm\frac{5}{\sqrt{6}}\)
Với \(k=\frac{5}{\sqrt{6}}\)thì \(\hept{\begin{cases}a=\frac{15}{\sqrt{6}}\\b=\frac{20}{\sqrt{6}}\\c=\frac{25}{\sqrt{6}}\end{cases}}\)
Với \(k=-\frac{5}{\sqrt{6}}\)thì \(\hept{\begin{cases}a=-\frac{15}{\sqrt{6}}\\b=-\frac{20}{\sqrt{6}}\\c=-\frac{25}{\sqrt{6}}\end{cases}}\)
Vậy ta có 2 bộ số sau \(\left\{\frac{15}{\sqrt{6}};\frac{20}{\sqrt{6}};\frac{25}{\sqrt{6}}\right\};\left\{-\frac{15}{\sqrt{6}};-\frac{20}{\sqrt{6}};-\frac{25}{\sqrt{6}}\right\}\)
