Giải:
Đặt \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=k\Rightarrow\left\{{}\begin{matrix}x=3k\\y=4k\\z=5k\end{matrix}\right.\)
Ta co: \(2x^2+2y^2-3z^2=-100\)
\(\Rightarrow18k^2+32k^2-75k^2=-100\)
\(\Rightarrow-25k^2=-100\)
\(\Rightarrow k^2=4\)
\(\Rightarrow k=\pm2\)
+) \(k=2\Rightarrow x=6,y=8,z=10\)
+) \(k=-2\Rightarrow x=-6,y=-8,z=-10\)
Vậy bộ số \(\left(x;y;z\right)\) là \(\left(6;8;10\right);\left(-6;-8;-10\right)\)
Ta có: \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\) => \(\dfrac{2x^2}{18}=\dfrac{2y^2}{32}=\dfrac{3z^2}{75}\)
Áp dụng tính chất của dãy tỉ số = nhau ta có:
\(\dfrac{2x^2}{18}=\dfrac{2y^2}{32}=\dfrac{3z^2}{75}\) = \(\dfrac{2x^2+2y^2-3z^2}{18+32-75}\) = \(\dfrac{-100}{-25}\) = 4
=> \(\left\{{}\begin{matrix}2x^2=72\\2y^2=128\\3z^2=300\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x^2=36\\y^2=64\\z^2=100\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=\pm6\\y=\pm8\\z=\pm10\end{matrix}\right.\)
Vì x,y,z cùng dấu => (x;y;z)= (6;8;10); (-6;-8;-10)
\(x=\dfrac{3z}{5};y=\dfrac{4z}{5}\)
=>\(2\left(\dfrac{3z}{5}\right)^2+2\left(\dfrac{4z}{5}\right)^2-3z^2=-100\)
=>\(\left(\dfrac{18}{25}+\dfrac{32}{25}-3\right)z^2=-100\)
\(z^2=100\Leftrightarrow z=\pm10\)
+ z =10 => x =3/5 .10 = 6 ; y =4/5 .10=8
+z =-10 => x = -6 ; y = -8
vậy ( x;y;z) là ( 6;8;10) hoặc (-6;-8;-10)
Từ \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\Rightarrow\dfrac{x^2}{9}=\dfrac{y^2}{16}=\dfrac{z^2}{25}\Rightarrow\dfrac{2x^2}{18}=\dfrac{2y^2}{32}=\dfrac{3z^2}{75}\)
Áp dụng tc dãy tỉ số "=" nhau ta có:
\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{2x^2}{18}=\dfrac{2y^2}{32}=\dfrac{3z^2}{75}=\dfrac{2x^2+2y^2-3z^2}{18+32-75}=\dfrac{-100}{-25}=4\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=4\Rightarrow x=4\cdot3=12\\\dfrac{y}{4}=4\Rightarrow y=4\cdot4=16\\\dfrac{z}{5}=4\Rightarrow z=4\cdot5=20\end{matrix}\right.\)
