\(\dfrac{x}{4}=\dfrac{y}{4}=\dfrac{z}{5}=>\dfrac{2x^2}{32}=\dfrac{2y^2}{32}=\dfrac{3z^2}{75}\)
AD t/c của dãy tỉ số bằng nhâu ta có
\(\dfrac{2x^2}{32}=\dfrac{2y^2}{32}=\dfrac{3z^2}{75}=\dfrac{2x^2+2y^2-3z^2}{32+32-75}=\dfrac{-100}{-11}=\dfrac{100}{11}\)
\(=>\left[{}\begin{matrix}x=\dfrac{400}{11}\\y=\dfrac{400}{11}\\z=\dfrac{500}{11}\end{matrix}\right.\)