Ta có: +) \(\dfrac{x}{2}=\dfrac{y}{3}=>\dfrac{x}{8}=\dfrac{y}{12}\)
+) \(\dfrac{y}{4}=\dfrac{z}{5}=>\dfrac{y}{12}=\dfrac{z}{15}\)
Từ trên suy ra \(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x^2-y^2}{64-144}=\dfrac{-16}{-80}=\dfrac{1}{5}\)
=> x=\(\dfrac{1}{5}.8=\dfrac{8}{5}\) ; y=\(\dfrac{1}{5}.12=\dfrac{12}{5}\) ; z=\(\dfrac{1}{5}.15=\dfrac{15}{5}=3\)
Vậy x=\(\dfrac{8}{5}\) ; y=\(\dfrac{12}{5}\) ; z=3