Theo đề ta có:
\(x:y:z=3:4:5\) và \(2x^2+2y^2-3z^2=-100\)
từ \(x:y:z=3:4:5\Rightarrow\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\)
Áp dụng tính chất dãy tỉ số băng nhau ta có:
\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{2x^2}{2.3^2}=\dfrac{2y^2}{2.4^2}=\dfrac{3z^2}{3.5^2}=\dfrac{2x^2+2y^2-3z^2}{18+32-75}=\dfrac{-100}{-25}=4\)
\(\dfrac{x}{3}=4\Rightarrow x=4.3=12\)
\(\dfrac{y}{4}=4\Rightarrow y=4.4=16\)
\(\dfrac{z}{5}=4\Rightarrow z=4.5=20\)
Vậy x=12 ; y=16 ; z=20
Ta có :
\(2x^2+2y^2-3z^2=-100\)
\(x:y:z=3:4:5\)
\(\Leftrightarrow\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\)
\(\Leftrightarrow\dfrac{x^2}{9}=\dfrac{y^2}{16}=\dfrac{z^2}{25}\)
\(\Leftrightarrow\dfrac{2x^2}{18}=\dfrac{2y^2}{32}=\dfrac{3z^2}{75}\)
Áp dụng t/c dãy tỉ số bằng nhau ta có ;
\(\dfrac{2x^2}{18}=\dfrac{2y^2}{32}=\dfrac{3x^2}{75}=\dfrac{2x^2+2y^2-3z^2}{18+32-75}=\dfrac{-100}{-25}=4\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=4\Leftrightarrow x=12\\\dfrac{y}{4}=4\Leftrightarrow y=16\\\dfrac{z}{5}=4\Leftrightarrow z=20\end{matrix}\right.\)
Vậy ..
