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

Suy ra

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

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

\(x+z+2=2y\)(3)

\(x+y-3=2z\)(4)

(2)-(1) ta có

\(1-x=2x-\frac{1}{2}\Rightarrow3x=\frac{3}{2}\Rightarrow x=\frac{1}{2}\)

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

\(y=-z\)

\(x+z+2=\frac{1}{2}+2-y==\frac{5}{2}-y\)

\(\frac{\frac{5}{2}-y}{y}=\frac{5}{2y}-1=2\Leftrightarrow\frac{5}{2y}=3\Leftrightarrow y=\frac{5}{6}\)

\(z=-\frac{5}{6}\)

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

Áp dụng TC DTSBN ta có :

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

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

\(\Leftrightarrow\hept{\begin{cases}x+y+z+1=3x\\x+y+z+1=3y\\x+y+z-2=3z\end{cases}}\) và \(x+y+z=\frac{1}{2}\)

\(\Leftrightarrow\hept{\begin{cases}\frac{1}{2}+1=3x\\\frac{1}{2}+1=3y\\\frac{1}{2}-2=3z\end{cases}}\)

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

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

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

\(\frac{y+z+1+x+z+1+x+y-3}{x+y+z}\)=\(\frac{2\left(X+Y+Z\right)}{x+y+z}\)=2  =>x+y+z=\(\frac{1}{2}\)   tu lam di nhe

Áp dụng t/c dãy tỉ số bằng nhau:

\(\frac{x+y+2019}{z}=\frac{y+z-2020}{x}=\frac{z+x+1}{y}=\frac{2}{x+y+z}\)

\(=\frac{x+y+2019+y+z-2020+z+x+1}{z+x+y}=2\)

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

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

Thay vào đầu bài:

\(\frac{1-z+2019}{z}=\frac{1-x-2020}{x}=\frac{1-y+1}{y}\)

\(\Leftrightarrow\frac{2020-z}{z}=\frac{-2019-x}{x}=\frac{2-y}{y}\)

\(\Leftrightarrow\frac{2020}{z}=\frac{-2019}{x}=\frac{2}{y}=\frac{2020-2019+2}{x+y+z}=3\)(Theo t/c dãy tỉ số bằng nhau)

\(\Rightarrow\hept{\begin{cases}z=\frac{2020}{3}\\x=\frac{-2019}{3}\\y=\frac{2}{3}\end{cases}}\)

ĐK: x , y, z, x+y+z khác 0

Áp dụng dãy tỉ số bằng nhau: ( kiến thức trong SGK lớp 7 em tìm hiểu lại nhé! )

\(\frac{x+y+2019}{z}=\frac{y+z-2020}{x}=\frac{z+x+1}{y}=\frac{x+y+2019+y+z-2020+z+x+1}{x+y+z}\)

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

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

  \(\frac{x+y+2019}{z}=2\Leftrightarrow x+y+2019=2z\)(2)

\(\frac{y+z-2020}{x}=2\Leftrightarrow y+z-2020=2x\) (3)

\(\frac{z+x+1}{y}=2\Leftrightarrow z+x+1=2y\) (4)

Từ (1) <=> x + y = 1 - z ; y +z =1 - x ; z + x = 1 -y . Lần lượt thế vào (2) ; (3) ; (4) để tìm x, y, z