Áp dụng tính chất dãy tie số bằng nhau ta có:

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

\(\Rightarrow\hept{\begin{cases}x-y-z=-x\\y-z-x=-y\\z-y-x=-z\end{cases}\Rightarrow\hept{\begin{cases}y+z=-2x\\z+x=-2y\\x+y=-2z\end{cases}}}\)

\(\Rightarrow\left(1+\frac{y}{x}\right)\left(1+\frac{z}{y}\right)\left(1+\frac{x}{z}\right)=\frac{\left(x+y\right)}{x}.\frac{\left(y+z\right)}{y}.\frac{\left(z+x\right)}{z}=-\frac{8xyz}{xyz}=-8\)

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=(-zyx)/(xyz)

B=-1

\(x+y+z=0\)

⇔\(-x=y+z\)

⇔\(x^2=\left(y+z\right)^2\) 

⇔\(x^2=y^2+2yz+z^2\) 

⇔\(y^2+z^2-x^2=-2yz\)

Tương tự:

\(z^2+x^2-y^2=-2zx\)

\(x^2+y^2-z^2=-2xy\)

➞ S = \(\dfrac{1}{-2xy}+\dfrac{1}{-2yz}+\dfrac{1}{-2zx}=\dfrac{x+y+z}{-2xyz}=0\) 

Vậy S = 0

Ta có:

\(x+y+z=0\)

\(\Rightarrow\left(x+y\right)^2=\left(-z\right)^2\)

\(\Rightarrow x^2+y^2+2xy=z^2\)

\(\Rightarrow x^2+y^2-z^2=-2xy\)

Tương tự ta được:
\(S=\frac{1}{-2xy}+\frac{1}{-2yz}+\frac{1}{-2zx}=-\frac{1}{2}\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)=-\frac{1}{2}\cdot\frac{x+y+z}{xyz}=0\)

Vậy S=0

ta có : x - y - z = 0   =>   \(\hept{\begin{cases}x=y+z\\y=x-z\\z=x-y\end{cases}}\)    =>  \(\hept{\begin{cases}x=y+z\\y=x-z\\-z=y-x\end{cases}}\)

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

x-y-z=0 ta có x-z=y,y-x=-z,y+z=x

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

thay vào được: y/x.-z/y.x/z=-1

cảm ơn các bạn nha

1/x+1/y-1/z=(yz+xz-xy)/(xyz)=0 vì x,y,z#0 =>yz+xz-xy=0

x^2 + y^2 +z^2=(x+y-z)^2 +2(xz+yz-xy)=4

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