gấp vãi !!!!!!!!!

a.

\(\left(xy\right).\left(yz\right).\left(xz\right)=3.6.18\\ \Rightarrow\left(x.y.z\right)^2=324\\ \Rightarrow\left[{}\begin{matrix}x.y.z=18\\x.y.z=-18\end{matrix}\right.\)

Nếu x.y.z=18

\(\Rightarrow\left[{}\begin{matrix}z=\left(xyz\right):\left(xy\right)=18:3=6\\x=\left(xyz\right):\left(yz\right)=18:6=3\\y=\left(xyz\right):\left(xz\right)=18:18=1\end{matrix}\right.\)

Nếu x.y.z = -18

\(\Rightarrow\left\{{}\begin{matrix}z=-6\\x=-3\\y=-1\end{matrix}\right.\)

Vậy...

a)Ta có:xy.yz.xz=3.6.18

x^2.y^2.z^2=324

(x.y.z)^2=18^2

x.y.z=18

Do đó:x=18:6=3

y=18:18=1

z=18:3=6

b)Ta có:xy.yz.xz=1.8.18

x^2.y^2.z62=144

(x.y.z)^2=12^2

x.y.z=12

Do đó:x=12:8=1,5

y=12:18=2/3

z=12:1=12

Lấy xy*yz*xz=(xyz)^2=36

xyz=6

x=xyz/yz=6/6=1

y=xy/x=2/1=2

z=yz/y=6/2=3

con TH xyz=-6 giai tuong tu nhe

Ta có xy=2 ;yz=6;zx=3

=> xy.yz.zx=2.6.3 => (xyz)^2=36

*xyz=6 => z=3;x=1;y=2

*xyz=-6 => z=-3;x=-1 ;y=-2

Ta có: \(xy=2\Rightarrow y=\frac{2}{x}\left(1\right)\)

         \(yz=6\Rightarrow y=\frac{6}{z}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\frac{2}{x}=\frac{6}{z}\)

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

Đặt \(\frac{x}{2}=\frac{z}{6}=k\Rightarrow\hept{\begin{cases}x=2k\\z=6k\end{cases}}\)

\(xz=3\Rightarrow2k\cdot6k=3\)

\(\Rightarrow12\cdot k^2=3\Rightarrow k^2=0,25\)

\(\Rightarrow k=0,5\)

\(x=2k\Rightarrow x=2\cdot0,5=1\)

\(z=6k\Rightarrow z=6\cdot0,5=3\)

Mà y=2/x => y=2:1=2

Vậy x=1; y=2;z=3

\(A=\frac{x}{xy+x+1}+\frac{y}{yz+y+1}+\frac{z}{zx+z+1}\)

\(A=\frac{xz}{xyz+xz+z}+\frac{yxz}{yz.xz+xyz+xz}+\frac{z}{zx+z+1}\) Thay xyz=1 vào ta được:

\(A=\frac{xz}{xz+z+1}+\frac{1}{z+1+xz}+\frac{z}{zx+z+1}\)

\(A=\frac{zx+z+1}{zx+z+1}=1\)

=> A=1