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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
\(A=\left(1-\frac{z}{x}\right)\left(1-\frac{x}{y}\right)\left(1+\frac{y}{z}\right)=\frac{\left(x-z\right)\left(y-x\right)\left(y+z\right)}{xyz}=\frac{y.\left(-z\right).x}{xyz}=-1\)
x - y - z = 0
x = y + z
y = x - z
z = x - y => -z = y - x
B = (1 - z/x)(1 - x/y) (1 + y/z)
B = (x/x - z/x)( y/y - x/y) ( z/z + y/z)
B = \(\frac{x-z}{x}\cdot\frac{y-x}{y}\cdot\frac{z+x}{z}=\frac{y}{x}\cdot\frac{-z}{y}\cdot\frac{x}{z}=-1\)
\(\dfrac{y+z-x}{x}=\dfrac{z+x-y}{y}=\dfrac{x+y-z}{z}\\ \Rightarrow\dfrac{y+z-x}{x}+2=\dfrac{z+x-y}{y}+2=\dfrac{x+y-z}{z}+2\\ \Rightarrow\dfrac{x+y+z}{x}=\dfrac{x+y+z}{y}=\dfrac{x+y+z}{z}\\ \Rightarrow x=y=z\\ \Rightarrow A=\left(1+1\right).\left(1+1\right).\left(1+1\right)=8\)
\(y=x+z\Rightarrow\hept{\begin{cases}x-y=-z\\y-z=x\end{cases}}\)
Ta ó: \(M=\left(1+\frac{x}{z}\right)\left(1-\frac{y}{x}\right)\left(1-\frac{z}{y}\right)\)
\(=\frac{x+z}{z}\cdot\frac{x-y}{y}\cdot\frac{y-z}{y}\)
\(=\frac{y}{z}\cdot\frac{-z}{x}\cdot\frac{x}{y}=-1\)
Vậy M=-1