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

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

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x+y+z=o suy ra x+y=-z ;z+x=-y;z+y=-x

B=-z.-x.-y=-xyz=-(-2016)=2016

Từ giả thiết x + y + z = 0 => x + y = - z; y + z = -x và x + z = -y

Vậy B = (x + y)(y + z) (z + x) = -z.(-x).(-y) = - xyz = 2016

Vậy B = 2016

Từ giả thiết x + y + z = 0 => x + y = - z; y + z = -x và x + z = -y

Vậy B = (x + y)(y + z) (z + x) = -z.(-x).(-y) = - xyz = 2016

Vậy B = 2016

Từ x+y+z=0

=>x+y=-z

x+z=-y

y+z=-x

thay vào C,ta có:C=(x+y)(y+z)(x+z)

=(-z)(-x)(-y)

Vì x.y.z=-2017

=>(-z)(-x)(-y)=2017

Vậy C=(x+y)(y+z)(x+z)=2017

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

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

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

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

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

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

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

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

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

\(=1\)