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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
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 = \(\frac{x-z}{x}\cdot\frac{y-x}{y}\cdot\frac{z+x}{z}=\frac{y}{x}\cdot\frac{-z}{y}\cdot\frac{x}{z}=-1\)
\(A=\left(1-\frac{z}{x}\right)\left(1-\frac{x}{y}\right)\left(1+\frac{y}{z}\right)=\frac{\left(x-z\right)\left(y-x\right)\left(y+z\right)}{xyz}=\frac{y.\left(-z\right).x}{xyz}=-1\)
Ta có: x-y-z = 0
\(\Rightarrow\) x = y+z
\(\Rightarrow\)y = x-z
\(\Rightarrow\)z = x-y
Thay vào B ta suy ra: \(\left(1-\frac{z}{x}\right)\left(1-\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\)
= \(\left(1-\frac{x-y}{x}\right)\left(1-\frac{y+z}{y}\right)\left(1+\frac{x-z}{z}\right)\)
= \(\left(\frac{-y}{x}\right).\left(\frac{z}{y}\right).\left(\frac{x}{z}\right)\)
= -y/y
= -1
Vậy B = -1
Ta có : \(A=\left(1-\frac{z}{x}\right)\left(1+\frac{x}{y}\right)\left(1-\frac{y}{z}\right)=\frac{x-z}{x}\cdot\frac{x+y}{y}\cdot\frac{z-y}{z}\)
\(x+y-z=0\Leftrightarrow\hept{\begin{cases}x+y=z\\x-z=-y\\z-y=x\end{cases}}\) thay vào A ta được :
\(A=\frac{-y}{x}\cdot\frac{z}{y}\cdot\frac{x}{z}==\frac{-y.z.x}{x.y.z}=-1\)
Ta có : \(B=\frac{x+y}{y}.\frac{z+y}{z}=\frac{x+z}{x}=\frac{\left(x+y\right)\left(z+y\right)\left(x+z\right)}{xyz}\)
Từ \(\frac{y+z-x}{x}=\frac{z+x-y}{y}=\frac{x+y-z}{z}\)
\(\Rightarrow\frac{y+z-x}{x}+2=\frac{z+x-y}{y}+2=\frac{x+y-z}{z}+2\)
\(\Rightarrow\frac{x+y+z}{x}=\frac{x+y+z}{y}=\frac{x+y+z}{z}\)
Nếu x + y + z = 0
=> x + y = - z
=> z + y = - x
=> z + x = - y
Khi đó : B = \(\frac{\left(-x\right)\left(-y\right)\left(-z\right)}{xyz}=-\frac{xyz}{xyz}=-1\)
Nếu x + y + z \(\ne\)0
=> \(\frac{1}{x}=\frac{1}{y}=\frac{1}{z}\Rightarrow x=y=z\)
Khi đó \(B=\frac{\left(x+y\right)^3}{x^3}=\frac{\left(2x\right)^3}{x^3}=\frac{2^3.x^3}{x^3}=8\)
Vậy nếu x + y + z = 0 B = - 1
nếu x + y + z \(\ne\)0 thì B = 8
