theo bài ra ta có \(\frac{x^3}{8}\)=\(\frac{y^3}{64}\)=\(\frac{z^3}{126}\)=>\(\frac{x^2}{4}\)=\(\frac{y^2}{16}\)=\(\frac{z^2}{36}\) và \(x^2\)+\(y^2\)+\(z^2\)=19
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\frac{x^2}{4}\)=\(\frac{y^2}{16}\)=\(\frac{z^2}{36}\)=\(\frac{x^2+y^2+z^2}{4+16+36}\)=\(\frac{19}{56}\)
lúc đó: \(\frac{x^2}{4}\)=\(\frac{19}{56}\)=>\(x^2\)=\(\frac{19}{14}\)=>x=\(\pm\sqrt{\frac{19}{14}}\)
\(\frac{y^2}{16}\)=\(\frac{19}{56}\)=>\(y^2\)=\(\frac{38}{7}\)=>\(\pm\sqrt{\frac{38}{7}}\)
\(\frac{z^2}{36}\)=\(\frac{19}{56}\)=>\(z^2\)=\(\frac{171}{14}\)=>\(\pm\sqrt{\frac{171}{14}}\)
vậy \(\left\{{}\begin{matrix}x=\frac{19}{14}\\y=\frac{38}{7}\\z=\frac{171}{14}\end{matrix}\right.\)hoặc \(\left\{{}\begin{matrix}x=-\frac{19}{14}\\y=-\frac{38}{7}\\z=-\frac{171}{14}\end{matrix}\right.\)
