Áp dụng tích chất dãy tỉ số bằng nhau ta có: 

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

=> x=y=z 

Ta có: 1 + x/y = (x+y)/y = (y+y)/y = 2y/y = 2

          1+ y/z = (y+z)/z = (z+z)/z = 2z/z = 2

    1 + z/x = (z+x)/z = (x+x)/x = 2x/x = 2

Vậy B= 2.2.2 = 8

z khác 0 thỏa mãn điều kiện $\frac{y+z-x}{x - Giúp tôi giải ...

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 

chỉ có lm thì mới có ăn

Cộng vế 2 đẳng thức đầu lại ta được 

(y+z-x+z+x-y+z+y-z)/(x+y+z)=2 nên (x+z-y)/y=2 hay x+z=3y, tương tự y+z=3x, x+y=3z nên GT=27

theo t/c dãy tỉ số=nhau:

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

=>x=y=z

\(1+\frac{x}{y}=\frac{x+y}{y}=\frac{y+y}{y}=\frac{2y}{y}=2\)

\(1+\frac{y}{z}=\frac{y+z}{z}=\frac{z+z}{z}=\frac{2z}{z}=2\)

\(1+\frac{z}{x}=\frac{z+x}{x}=\frac{x+x}{x}=\frac{2x}{x}=2\)

=>B=2.2.2=8

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

\(\Leftrightarrow x=\frac{1}{2};y=\frac{1}{2};z=-\frac{1}{2}\)

\(\Leftrightarrow B=\left(1+\frac{\frac{1}{2}}{\frac{1}{2}}\right)\left(1+\frac{\frac{1}{2}}{\frac{-1}{2}}\right)\left(1+\frac{\frac{-1}{2}}{\frac{1}{2}}\right)\)=0

cộng thêm 2 mỗi bên : \(\frac{y+z-x}{x}+2=\frac{z+x-y}{y}+2=\frac{x+y-z}{z}+2\)

\(\frac{y+z+x}{x}=\frac{z+x+y}{y}=\frac{x+y+z}{z}\) => x =y =z  ( vì tử = nhau)

=> B = 2.2.2 =8

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{2013}=\frac{1}{x+y+z}\Rightarrow\frac{yz+xz+xy}{xyz}=\frac{1}{x+y+z}\Rightarrow\left(yz+xz+xy\right)\left(x+y+z\right)=xyz\)

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

\(\Rightarrow y^2z+yz^2+x^2z+xz^2+x^2y+xy^2+2xyz=0\)

\(\Rightarrow\left(x^2y+x^2z+xy^2+xyz\right)+\left(y^2z+xz^2+y^2z+xyz\right)=0\)

\(\Rightarrow x\left(xy+xz+y^2+yz\right)+z\left(yz+xz+y^2+xy\right)=0\)

\(\Rightarrow\left(x+z\right)\left(xy+xz+y^2+yz\right)=\left(x+z\right)\left(x\left(y+z\right)+y\left(y+z\right)\right)=\left(x+y\right)\left(y+z\right)\left(x+z\right)=0\)

\(\Rightarrow\hept{\begin{cases}x+y=0\Rightarrow x^3+y^3=0\\y+z=0\Rightarrow y^5+z^5=0\\x+z=0\Rightarrow z^7+x^7=0\end{cases}}\)

\(\Rightarrow A=\left(x^3+y^3\right)\left(y^5+z^5\right)\left(z^7+x^7\right)=0\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

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

--> y + z - x = x; z + x - y = y; x + y - z = z

--> y + z = 2x; z + x = 2y; x + y = 2z

Ta có: 

B = (x + y)/y.(y + z)/z.(z + x)/x

= 2z/y.2x/z.2y/x = 8