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

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

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

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

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    ^_^

\(\left(x-1\right)^2\ge0\Rightarrow x^2-2x+1\ge0\Rightarrow x^2+1\ge2x\)

\(\left(y-2\right)^2\ge0\Rightarrow y^2-4y+4\ge0\Rightarrow y^2+4\ge4y\)

\(\left(z-3\right)^2\ge0\Rightarrow z^2-6z+9\ge0\Rightarrow z^2+9\ge6z\)

Do đó: \(\left(x^2+1\right)\left(y^2+4\right)\left(z^2+9\right)\ge2x.4y.6z=48xyz\)

Dấu "=" xảy ra khI: \(\hept{\begin{cases}x-1=0\\y-2=0\\z-3=0\end{cases}\Rightarrow\hept{\begin{cases}x=1\\y=2\\z=3\end{cases}}}\)

Vậy \(C=\frac{1^3+2^3+3^3}{\left(1+2+3\right)^3}=\frac{6^2}{6^3}=\frac{1}{6}\)

Chúc bạn học tốt.

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