Đáp án:
$x.y.z=\pm480$
Giải thích các bước giải:
Theo tính chất dãy tỉ số bằng nhau ta có:
$\dfrac{3z-4y}{5}=\dfrac{5y-3x}{4}=\dfrac{4x-5z}{3}=\dfrac{5(3z-4y)+4(5y-3x)+3(4x-5z)}{5.5+4.4+2.2}=\dfrac{15z-20y+20y-12x+12x-15z}{50}=\dfrac{0}{50}=0$
$\Rightarrow \left\{\begin{array}{l}\dfrac{3z-4y}{5}=0\\ \dfrac{5y-3x}{4}=0 \\ \dfrac{4x-5z}{3}=0\end{array} \right.\Rightarrow \left\{\begin{array}{l}3z-4y=0\\ 5y-3x=0 \\ 4x-5z=0\end{array} \right.$
$\Rightarrow \left\{\begin{array}{l}3z=4y\\ 5y=3x \\4x=5z\end{array} \right.\Rightarrow \left\{\begin{array}{l}\dfrac{z}{4}=\dfrac{y}{3}\\ \dfrac{y}{3}=\dfrac{x}{5} \\ \dfrac{x}{5}=\dfrac{z}{4}\end{array} \right.$
$\Rightarrow \dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{4}$
Theo tính chất dãy tỉ số bằng nhau ta có:
$\Rightarrow\dfrac{x^2}{25}=\dfrac{y^2}{9}=\dfrac{z^2}{16}= \dfrac{x^2-z^2}{25-16}=\dfrac{36}{9}=4$
$\Rightarrow \left\{\begin{array}{l}\dfrac{x^2}{25}=4\\ \dfrac{y^2}{16}=4\\ \dfrac{z^2}{9}=4\end{array} \right.\Rightarrow \left\{\begin{array}{l}x^2=100\\ y^2=36 \\ z^2=64\end{array} \right.$
$\Rightarrow x^2.y^2.z^2=100.36.64$
$\Rightarrow (x.y.z)^2=480^2$
$\Rightarrow |x.y.z|=480$
$\Rightarrow x.y.z=\pm480$
