Thèo đề bài, ta có:

\(\frac{x^3}{2^3}=\frac{y^3}{4^3}=\frac{z^3}{6^3}=\frac{x}{2}=\frac{y}{4}=\frac{z}{6}=\frac{x^2}{4}=\frac{y^2}{16}=\frac{z^2}{36}=\frac{x^2+y^2+z^2}{4+16+36}=\frac{14}{56}=\frac{1}{4}\)

x ; y ; z thì bạn tự tìm nhé , chắc cái này không khó đâu nhỉ ??

\(\frac{x^3}{8}=\frac{y^3}{64}=\frac{z^3}{216}\Rightarrow\frac{x}{2}=\frac{y}{4}=\frac{z}{6}\Rightarrow\frac{x^2}{4}=\frac{y^2}{16}=\frac{z^2}{36}\) \(=\frac{x^2+y^2+z^2}{4+16+36}=\frac{14}{56}=\frac{1}{4}\)

\(\frac{x}{2}=\frac{1}{4}\Rightarrow x=\frac{1}{2}\)

\(\frac{y}{4}=\frac{1}{4}\Rightarrow y=1\)

\(\frac{z}{6}=\frac{1}{4}\Rightarrow z=\frac{3}{2}\)

\(\frac{x^3}{8}=\frac{y^3}{64}=\frac{z^3}{216}\Leftrightarrow\frac{x^3}{2^3}=\frac{y^3}{4^3}=\frac{z^3}{6^3}\Leftrightarrow\frac{x}{2}=\frac{y}{4}=\frac{z}{3}\)

Đến đây tự làm được rồi nhé !    

=>\(\frac{x^3}{2^3}=\frac{y^3}{4^3}=\frac{z^3}{6^3}\)=>\(\frac{x}{2}=\frac{y}{4}=\frac{z}{6}\)=>\(\frac{x^2}{2^2}=\frac{y^2}{4^2}=\frac{z^2}{6^2}\)

Ap dụng tính chất dãy tỉ số bằng nhau ta có:

\(\frac{x^2}{2^2}=\frac{y^2}{4^2}=\frac{z^2}{6^2}=\frac{x^2+y^2+z^2}{2^2+4^2+6^2}=\frac{14}{56}=\frac{1}{4}\)(Vì x²+y²+z²=14)

=>\(\frac{x^2}{2^2}=\frac{1}{4}=>x^2=1=>x^2=1;x=-1\)

=>\(\frac{y^2}{4^2}=\frac{1}{4}=>y^2=4=>y=2;y=-2\)

=>\(\frac{z^2}{6^2}=\frac{1}{4}=>z^2=9=.z=3;z=-3\)

Vậy x=1 ; y=2 ; z=3  hoặc x=-1 ; y=-2 ; z=-3

Hình như đề là:\(x^2+y^2+z^{2=14}\)  mới đúng

\(\frac{x^3}{8}=\frac{y^3}{64}=\frac{z^3}{216}=\frac{x^3}{2^3}=\frac{y^3}{4^3}=\frac{z^3}{6^3}\)

\(\Rightarrow\frac{x^2}{2^2}=\frac{y^2}{4^2}=\frac{z}{6}\)

Theo tính chất dãy tỉ số bằng nhau ta có :

\(\frac{x^2+y^2+z}{4+16+6}=\frac{14}{26}=\frac{7}{13}\)

\(\hept{\begin{cases}\frac{x^2}{4}=\frac{7}{13}\Rightarrow x=\sqrt{\frac{28}{13}}\\\frac{y^2}{16}=\frac{7}{13}\Rightarrow y=\sqrt{\frac{112}{13}}\\\frac{z}{6}=\frac{7}{13}\Rightarrow z=\frac{42}{13}\end{cases}}\)

Vậy ....

Ta có:\(\frac{x^3}{8}=\frac{y^3}{64}=\frac{z^3}{216}\Rightarrow\frac{x}{2}=\frac{y}{4}=\frac{z}{6}\)

Đặt \(\frac{x}{2}=\frac{y}{4}=\frac{z}{6}=k.\Rightarrow x=2k;y=4k;z=6k\)

Mà\(x^2+y^2+z^2=14\)

\(\Rightarrow\left(2k\right)^2+\left(4k\right)^2+\left(6k\right)^2=14\)

\(\Rightarrow4.k^2+16.k^2+36.k^2=14\)

\(\Rightarrow k^2\left(4+16+36\right)=14\)

\(\Rightarrow k^2=14:56\)

\(\Rightarrow k^2=\frac{1}{4}\)

\(\Rightarrow k=\pm\frac{1}{2}\)

\(\Rightarrow x=\pm1,y=\pm2,z=\pm3\)

\(\frac{x^3}{8}=\frac{y^3}{64}=\frac{z^3}{216}\Rightarrow\frac{x}{2}=\frac{y}{4}=\frac{z}{6}\Rightarrow\frac{x^2}{4}=\frac{y^2}{16}=\frac{z^2}{36}=\frac{x^2+y^2+z^2}{4+16+36}=\frac{14}{56}=\frac{1}{4}\)

\(\Rightarrow\hept{\begin{cases}\frac{x^2}{4}=\frac{1}{4}\\\frac{y^2}{16}=\frac{1}{4}\\\frac{z^2}{36}=\frac{1}{4}\end{cases}\Rightarrow\hept{\begin{cases}x^2=1\\y^2=4\\z^2=9\end{cases}\Rightarrow}\hept{\begin{cases}x=\pm1\\y=\pm2\\z=\pm3\end{cases}}}\)

\(\frac{x^3}{8}=\frac{y^3}{64}=\frac{z^3}{216}\)=>\(\frac{x^3}{2^3}=\frac{y^3}{4^3}=\frac{z^3}{6^3}\)=>\(\frac{x}{2}=\frac{y}{4}=\frac{z}{6}\)

=>\(\frac{x^2}{2^2}=\frac{y^2}{4^2}=\frac{z^2}{6^2}\)=>\(\frac{x^2}{4}=\frac{y^2}{16}=\frac{z^2}{36}\)

Aps dụng t/c dãy tỉ số bằng nhau ta có:

\(\frac{x^2}{4}=\frac{y^2}{16}=\frac{z^2}{36}=\frac{x^2+y^2+z^2}{4+16+36}=\frac{14}{56}=\frac{1}{4}\)(vì x²+y²+z²=14)

=>\(\frac{x^2}{4}=\frac{1}{4}=>x^2=1=>x=1;x=-1\)

=>\(\frac{y^2}{16}=\frac{1}{4}=>y^2=4=>y=2;y=-2\)

=>\(\frac{z^2}{36}=\frac{1}{4}=>z^2=9=>z=3;z=-3\)

Vậy x=1; y=2 ; z=3

Hoặc x=-1 ;y=-2 ;z=-3

Ta có :\(\frac{x^3}{8}=\frac{y^3}{64}=\frac{z^2}{216}\Rightarrow\frac{x}{2}=\frac{y}{4}=\frac{z}{6}\Rightarrow\frac{x^2}{4}=\frac{y^2}{16}=\frac{z^2}{36}\) và \(x^2+y^2+z^2=14\)

Áp đụng tính chất dãy tỉ số bằng nhau :

\(\frac{x^2}{4}=\frac{y^2}{16}=\frac{z^2}{36}=\frac{x^2+y^2+z^2}{4+16+36}=\frac{14}{56}=\frac{1}{4}\approx0,25\)

Suy ra : \(\frac{x^2}{4}=0,25\Rightarrow x^2=0,25.4=1\Rightarrow x=1\)

\(\frac{y^2}{16}=0,25\Rightarrow y^2=0,25.16=4\Rightarrow y=2\)

\(\frac{z^2}{36}=0,25\Rightarrow z^2=0,25.36=9\Rightarrow z=3\)

Vậy :x=1 , y=2 và z=3