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

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

Nếu   \(x+y+z=0\)thì   \(\hept{\begin{cases}x+y=-z\\y+z=-x\\z+x=-y\end{cases}}\)

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

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

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

Nếu  \(x+y+z\ne0\)thì   \(\frac{x-y-z}{x}=\frac{-x+y-z}{y}=\frac{-x-y+z}{z}=-1\)

suy ra:   \(\frac{x-y-z}{x}=-1\)            \(\Rightarrow\)       \(x-y-z=-x\)          \(\Rightarrow\)     \(y+z=2x\)

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

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

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

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

\(=\frac{2z}{x}.\frac{2x}{y}.\frac{2y}{z}=8\)

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

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

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

A = -z/x . (-x)/y . (-y)/z = -1

+ Nếu x + y + z khác 0

x-y-z/x = -x+y-z/y = -x-y+z/z

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

<=> y+z/x = x+z/y = x+y/z

Áp dụng t/c của dãy tỉ số = nhau ta có:

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

A = (x+y)/x . (y+z)/y . (x+z)/z = 8

\(\Rightarrow A=2.\)

Bài này hình như lớp 7 đúng ko, nếu lớp 7 thì mk giải đc

\(\dfrac{y+z-x}{x}=\dfrac{z+x-y}{y}=\dfrac{x+y-z}{z}\\ \Rightarrow\dfrac{y+z-x}{x}+2=\dfrac{z+x-y}{y}+2=\dfrac{x+y-z}{z}+2\\ \Rightarrow\dfrac{x+y+z}{x}=\dfrac{x+y+z}{y}=\dfrac{x+y+z}{z}\\ \Rightarrow x=y=z\\ \Rightarrow A=\left(1+1\right).\left(1+1\right).\left(1+1\right)=8\)

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

\(\rightarrow\begin{cases}x-y-z=-x\\y-x-z=-y\\z-x-y=-z\end{cases}\)

\(\leftrightarrow\begin{cases}y+z=2x\\z+x=2y\\x+y=2z\end{cases}\)

\(A=\frac{x+y}{z}.\frac{y+z}{x}.\frac{z+x}{y}=8\)

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

A=6

\(\frac{x-y-z}{x}=1-\frac{y+z}{x}\) tương tự con khác

=> x=y=z

=> A=6

Ta có \(x-y-z=0\)

\(\Rightarrow\hept{\begin{cases}x-z=y\\y-x=-z\\z+y=x\end{cases}}\)( 1 )

Ta có:

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

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

Thay điều ( 1 ) vào biểu thức ta có:

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

\(\Rightarrow B=\frac{y}{x}.\frac{-z}{y}.\frac{x}{z}\)

\(\Rightarrow B=-1\)

Vậy B = -1 

Ta có:

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

1-x/y=y/y-x/y=(y-x)/y(2)

1+y/z=z/z+y/z=(y+z)/z(3)

Mà x-y-z=0( theo đề)

=>x-z=y(*)

 x-y=z=>y-x=-z ( số đối) (**)

y+z=x(***)

 Thay (*),(**),(***) lần lượt vào (1),(2),(3) ta đc:

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

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

 Vậy A=-1

Thanks

:)

[e]134[e]

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

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

Do \(x-y-z=0\)

\(\Rightarrow x-z=y;y-x=-z;y+z=x\)

Khi đó \(A=\frac{y}{x}\cdot\frac{-z}{y}\cdot\frac{x}{z}=-1\)

Vậy A=-1

\(\frac{1}{xy+x+1}+\frac{y}{yz+y+1}+\frac{1}{xyz+yz+y}\)

\(=\frac{1}{xy+x+1}+\frac{y}{yz+y+1}+\frac{1}{1+yz+y}\)

\(=\frac{1}{xy+x+1}+\frac{y+1}{yz+y+1}\)

\(=\frac{yz}{xy\cdot yz+xyz+yz}+\frac{y+1}{yz+y+1}\)

\(=\frac{yz}{yz+y+1}+\frac{y+1}{yz+y+1}\)

\(=\frac{yz+y+1}{yz+y+1}\)

\(=1\)

\(\text{Ta có: }x-y-z=0\Rightarrow x=y+z\)

                                                  \(y=x-z\) 

                                                  \(z=x-y\)

\(\text{Mặt khác: }A=\left(1-\frac{z}{x}\right)\left(1-\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\)

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

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

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

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

                           \(=-1\)