Question

tìm x,y,z biết : x/ z+y+1=y/ x+z+1=z/ x+y-2 = x+y+z (x,y,z khác 0)

Answer 1

Ta có:

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

\(=\dfrac{x+y+z}{2\cdot\left(x+y+z\right)}=\dfrac{1}{2}\) \(\Rightarrow x+y+z=\dfrac{1}{2}\)

TH1 : x = y = z =0

TH2 : x;y;z ≠ 0

Ta có :

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

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

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

Mặt khác :

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

\(\Rightarrow2z+2+z=\dfrac{1}{2}\)

\(\Rightarrow3z+2=\dfrac{1}{2}\)

\(\Rightarrow3z=\dfrac{1}{2}-2\)

\(\Rightarrow3z=\dfrac{-3}{2}\)

\(\Rightarrow z=-\dfrac{1}{2}\)

\(\Rightarrow x=\dfrac{1}{2};y=\dfrac{1}{2};z=\dfrac{-1}{2}\)