x+y+z=0

=>x+y=-z

=>y+z=-x

=>z+x=-y

(1+x/y)(1+y/z)(1+z/x)

(y+x/y)(z+y/z)(x+z/x)

-z/y.-x/z.-y/x

=-1

\(A=\left(1-\frac{z}{x}\right)\left(1-\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\)

\(A=\frac{x-z}{x}\cdot\frac{y-x}{y}\cdot\frac{y+z}{z}\)

Do \(x-y-z=0\)

\(\Rightarrow x-z=y;y-x=-z;y+z=x\)

Khi đó \(A=\frac{y}{x}\cdot\frac{-z}{y}\cdot\frac{x}{z}=-1\)

Vậy A=-1

\(\frac{1}{xy+x+1}+\frac{y}{yz+y+1}+\frac{1}{xyz+yz+y}\)

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

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

\(=\frac{yz}{xy\cdot yz+xyz+yz}+\frac{y+1}{yz+y+1}\)

\(=\frac{yz}{yz+y+1}+\frac{y+1}{yz+y+1}\)

\(=\frac{yz+y+1}{yz+y+1}\)

\(=1\)

Ta có x+y+z = 0

\(\Rightarrow\left(+\right)x+y=-z\)

  \(\left(+\right)x+z=-y\)

\(\left(+\right)z+y=-x\)

C= (x+y)(y+z)(x+z)=(-z)(-x)(-y)=-1(xyz)=-1. 2= -2

\(x+y+z=0\Rightarrow\hept{\begin{cases}x+y=-z\\x+z=-y\\y+z=-x\end{cases}}\)

\(\Rightarrow\left(x+y\right)\left(y+z\right)\left(x+z\right)=-xyz=-2000\)

~~

x-y-z=0

=> x=y+z

     y=x-z

    -z=y-x

B=(1-z/x)(1-x/y)(1+y/z)

B=((x-z)/x)((y-x)/y)((z+y)/z)

B=(y/x)(-z/y)(x/z)

B=(-z.y.x)/(x.y.z)

B=-1

thank ban nha

B=-1    ^_^