Bài này ez thôi, làm mãi rồi.

Theo đề bài, ta có: \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)

=>\(\dfrac{xy+yz+xz}{xyz}=0\)

=> xy+yz+zx=0

=> \(\left\{{}\begin{matrix}xy=-yz-zx\\yz=-xy-zx\\zx=-xy-yz\end{matrix}\right.\)

Ta có: x²+2yz=x²+yz-xy-zx=(x-y)(x-z)

           y²+2xz=y²+xz-xy-yz=(x-y)(z-y)

           z²+2xy=z²+xy-yz-xz=(x-z)(y-z)

=> \(\dfrac{yz}{\left(x-y\right)\left(x-z\right)}+\dfrac{xz}{\left(x-y\right)\left(z-y\right)}+\dfrac{xy}{\left(x-z\right)\left(y-z\right)}=\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\dfrac{\left(x-y\right)\left(x-z\right)\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=1\)

x-y-z=0 => x=y+z

thế vào rồi tính B

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

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Rightarrow\frac{1}{x}+\frac{1}{y}=-\frac{1}{z};\frac{1}{x}+\frac{1}{z}=-\frac{1}{y};\frac{1}{y}+\frac{1}{z}=-\frac{1}{x}\)

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

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

\(=y\cdot-\frac{1}{y}+x\cdot-\frac{1}{x}+z\cdot-\frac{1}{z}=-1-1-1=-3\)

vậy A=-3

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    ^_^

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

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