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

x+y=2z

=> kz=2z

=>k=2

Theo tính chất dãy tỉ số bằng nhau ta có:

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

x+ y/z = 2 

2z = x + y

Vậy z = 2 

x+y+z=1?

Giải:

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

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

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

Xét \(\frac{x}{y+z+1}=\frac{1}{2}\)

\(\Rightarrow2x=y+z+1\)

\(\Rightarrow3x=x+y+z+1\)

\(\Rightarrow3x=1+1\)

\(\Rightarrow x=\frac{2}{3}\)

Xét \(\frac{y}{x+z+1}=\frac{1}{2}\)

\(\Rightarrow2y=x+z+1\)

\(\Rightarrow3y=x+y+z+1\)

\(\Rightarrow3y=1+1\)

\(\Rightarrow y=\frac{2}{3}\)

Xét \(\frac{z}{x+y-2}=\frac{1}{2}\)

\(\Rightarrow2z=x+y-2\)

\(\Rightarrow3z=x+y+z-2\)

\(\Rightarrow3z=1-2\)

\(\Rightarrow z=\frac{-1}{3}\)

Từ đó \(\frac{x+y}{z+1}=\frac{\frac{2}{3}+\frac{2}{3}}{\frac{-1}{3}+1}=\frac{\frac{4}{3}}{\frac{2}{3}}=\frac{4}{2}=2\)

Vậy \(\frac{x+y}{z+1}=2\)

Vì  ta có \(\frac{x+y}{z}=\frac{y+z}{x}=\frac{x+z}{y}\)và x+y = kz   =>  x=y=z => x+y = 2z . Mà x+y = kz = 2z => kz = 2z => k = 2

từ cái trên => x=y=z mà x+y=kz => k=2

k=2

tik cho mình nhé

ai tick mik đến 10 mik tick cho cả đời

cần mk giải chi tiết ko

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

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

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

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

theo t/c dãy t/s=nhau:

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

=>x+y=2z=kz(theo đề)

=>k=2

vậy k=2

Theo t/c dãy tỉ số = nhau:

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

=> \(\frac{x+y}{z}=2\Rightarrow x+y=2z=kz\Rightarrow k=2.\)

Vậy k=2.