Question

[Bảo Ngọc cute]

Tìm các số thực x,y,z biết

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

Answer 1

[Nguyễn Huy Tú]

Giải:

Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x+y-3}{z}=\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{1}{x+y+z}=\frac{x+y-3+y+z+1+x+z+2}{x+y+z}=\frac{2x+2y+2z}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)

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

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

Xét \(\frac{x+y-3}{z}=2\)

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

\(\Rightarrow x+y+z-3=3z\)

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

\(\Rightarrow\frac{-5}{2}=3z\)

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

Xét \(\frac{y+z+1}{x}=2\)

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

\(\Rightarrow x+y+z+1=3x\)

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

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

\(\Rightarrow x=\frac{1}{2}\)

Xét \(\frac{x+z+2}{y}=2\)

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

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

\(\Rightarrow\frac{1}{2}+2=3y\)

\(\Rightarrow\frac{5}{2}=3y\)

\(\Rightarrow y=\frac{5}{6}\)

Vậy bộ số \(\left(x;y;z\right)\) là \(\left(\frac{1}{2};\frac{5}{6};\frac{-5}{6}\right)\)