Theo đề ta có: xyz= 8.abc= xyz.abc= ax. by. cz= 8

                                                       hay ax.ax.ax= 8

=> (ax)³= 2³

=> ax= 2

Với ax= 2=> x= \(\frac{2}{a}\)

      by= 2=> y= \(\frac{2}{b}\)

      cz= 2=> z=\(\frac{2}{c}\)

Vậy x, y, z= \(\frac{2}{a},\frac{2}{b},\frac{2}{c}.\)

ta co : 

xyz=8/abc=>ax.by.cz=8 hay ax.ax.ax=8 <=> (ax)^3=2^3

=>ax=2=>x/a;y=2/b;z=2/c

x.y.z= 8/ a.b.c =>abc.xyz=8 
a.x=b.y=c.z =>(a.x)^3=(b.y)^3=(c.z)^3 =ax.by.cz=8 
* (a.x)^3 =8 =>a.x=2 =>x=2/a 
* (b.y)^3 =8 =>b.y=2 =>y=2/b 
* (c.z)^3 =8 =>c.z=2=>z=2/c

ta có: \(ax=by=cz\Rightarrow x:\frac{1}{a}=y:\frac{1}{b}=z:\frac{1}{c}=\frac{x}{\frac{1}{a}}=\frac{y}{\frac{1}{b}}=\frac{z}{\frac{1}{c}}=k.\)

\(\Rightarrow\hept{\begin{cases}x=\frac{k}{a}\\y=\frac{k}{b}\\z=\frac{k}{c}\end{cases}}\)

mà xyz = 8/abc \(\Rightarrow\frac{k}{a}\cdot\frac{k}{b}\cdot\frac{k}{c}=\frac{k^3}{abc}=\frac{8}{abc}\Rightarrow k^3=8=2^3\Rightarrow k=2\)

=> x = 2/a; y = 2/b; z = 2/c

ta co ax=by=cz 

suy ra x/(1/a)=y/(1/b)=z/(1/c)=k

suy ra x=k/a ; y=k/b ; z=k/c

ma xyz=8/abc suy ra k/a*k/b*k/c=k^3=8

suy ra k=2

suy ra x=2/a;y=2/b;z=2/c