Question

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

Answer 1

Lời giải:

Nếu $x+y+z=0$

$\Rightarrow \frac{x}{y+z+1}+\frac{y}{x+z+1}=\frac{z}{x+y-2}=x+y+z=0$

$\Rightarrow x=y=z=0$ (thỏa mãn)

Nếu $x+y+z\neq 0$. ADTCDTSBN ta có:

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

Ta có: $\frac{x}{y+z+1}=\frac{1}{2}$

$\Rightarrow 2x=y+z+1$. Mà $y+z=\frac{1}{2}-x$ nên:

$2x=\frac{1}{2}-x+1\Rightarrow x=\frac{1}{2}$

$\frac{y}{x+z+1}=\frac{1}{2}\Rightarrow 2y=x+z+1=\frac{1}{2}-y+1$

$\Rightarrow y=\frac{1}{2}$

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

Vậy $(x,y,z)=(0,0,0)$ hoặc $(\frac{1}{2}, \frac{1}{2}, \frac{-1}{2})$