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

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

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

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

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

Ta có : \(\frac{x+y}{z}=\frac{y+z}{x}=\frac{x+z}{y}\)       và   x + y = kz  (1)

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

=>\(\hept{\begin{cases}\frac{y+z}{x}=k\\\frac{x+z}{y}=k\end{cases}}\)=>\(\hept{\begin{cases}y+z=kx\left(2\right)\\x+z=ky\left(3\right)\end{cases}}\)

cộng vế với vế của (1),(2),(3) ,ta có :

2(x+y+z)=k(x+y+z)   => k=2

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

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 

chtt điĐặng Quốc Thắng