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

A=6

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

=> x=y=z

=> A=6

Dùng tính chất tỉ lệ thức:

\(\frac{x}{\left(y+z+1\right)}=\frac{y}{\left(x+z+1\right)}=\frac{z}{\left(x+y-2\right)}=0\Rightarrow x=y=z=0\) 

Áp dụng tính chất tỉ lệ thức: 

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

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

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

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

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

TA CÓ: \(\frac{x}{z+y+1}=\frac{y}{x+z+1}=\frac{z}{x+y-2}=\frac{x+y+z}{z+y+1+x+z+1+x+y-2}=\frac{1.\left(x+y+z\right)}{\left(1+1-2\right)+2x+2y+2z}\)

\(=\frac{1.\left(x+y+z\right)}{0+2.\left(x+y+z\right)}=\frac{1.\left(x+y+z\right)}{2.\left(x+y+z\right)}=\frac{1}{2}\)

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

\(\Rightarrow\frac{x}{z+y+1}=\frac{1}{2}\)\(\Rightarrow2x=z+y+1\)\(\Rightarrow3x=x+z+y+1\)\(\Rightarrow3x=\frac{1}{2}+1\Rightarrow3x=\frac{3}{2}\Rightarrow x=\frac{1}{2}\)

\(\frac{y}{x+z+1}=\frac{1}{2}\)\(\Rightarrow2y=x+z+1\Rightarrow3y=y+x+z+1\Rightarrow3y=\frac{1}{2}+1=\frac{3}{2}\Rightarrow y=\frac{1}{2}\)

\(\frac{z}{x+y-2}=\frac{1}{2}\)\(\Rightarrow2z=x+y-2\Rightarrow3z=x+y+z-2\Rightarrow3z=\frac{1}{2}-2=\frac{-3}{2}\Rightarrow z=\frac{-1}{2}\)

VẬY X= 1/2; Y= 1/2 ; Z= -1/2

CHÚC BN HỌC TỐT!!!!!!

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

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

áp dụng tính chất của dãy tỉ số bằng nhau ta có:\(\frac{ }{ }\)

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

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

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

=0+0+0+x+y+z/x+y+z=1

\(\Leftrightarrow\)x=y=z (*)

thay (*) vào B ta có:

B=(1+x/x)(1+x/x)(1+x/x)

  =2.2.2=8

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

\(...=\frac{y+z-x+z+x-y+x+y-z}{x+y+z}=\frac{x+y+z}{x+y+z}=1\)( vì x + y + z \(\ne\)0 )

\(\Rightarrow\hept{\begin{cases}\frac{y+z-x}{x}=1\\\frac{z+x-y}{y}=1\\\frac{x+y-z}{z}=1\end{cases}}\Rightarrow\hept{\begin{cases}y+z-x=x\\z+x-y=y\\x+y-z=z\end{cases}}\Rightarrow\hept{\begin{cases}y+z=2x\\z+x=2y\\x+y=2z\end{cases}}\Rightarrow x=y=z\)

Thế x = y = z vào B ta được :

\(B=\left(1+\frac{y}{y}\right)\left(1+\frac{x}{x}\right)\left(1+\frac{z}{z}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=2\cdot2\cdot2=8\)

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

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

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

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

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

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

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

=> \(\frac{y+z}{x}=\frac{z+x}{y}=\frac{x+y}{z}\)\(=\frac{y+z-z-x}{x-y}=\frac{y-x}{x-y}=-1\)(1)
Ta lại có \(B=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)=\frac{\left(x+y\right)\left(z+y\right)\left(x+z\right)}{xyz}\)(2)

Từ(1),(2) => \(B=-1.\left(-1\right).\left(-1\right)=-1\)

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

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

\(\Rightarrow\frac{y+z}{x}=\frac{z+x}{y}=\frac{x+y}{z}=\frac{y+z+z+x+x+y}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)( \(x,y,z\ne0\))

\(\Rightarrow y+z=2x\); \(z+x=2y\); \(x+y=2z\)(1)

Ta có: \(B=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)=\frac{x+y}{y}.\frac{y+z}{z}.\frac{x+z}{x}=\frac{\left(x+y\right)\left(y+z\right)\left(x+z\right)}{xyz}\)(2)

Từ (1) và (2) \(\Rightarrow B=\frac{2z.2x.2y}{xyz}=\frac{8xyz}{xyz}=8\)