Question

Tìm x,y,z
x+y+z=x/y+z-3=y/x+z-4=z/x+y+7

 

Answer 1

Lời giải:

Nếu $x+y+z=0$ thì:

$\frac{x}{y+z-3}=\frac{y}{x+z-4}=\frac{z}{x+y+7}=0$

$\Rightarrow x=y=z=0$

Nếu $x+y+z\neq 0$ thì, áp dụng TCDTSBN:

$x+y+z=\frac{x}{y+z-3}=\frac{y}{x+z-4}=\frac{z}{x+y+7}=\frac{x+y+z}{y+z-3+x+z-4+x+y+7}=\frac{x+y+z}{2(x+y+z)}=\frac{1}{2}$

$\Rightarrow \frac{y+z-3}{x}=\frac{x+z-4}{y}=\frac{x+y+7}{z}=2$

$\Rightarrow \frac{y+z-3}{x}+1=\frac{x+z-4}{y}+1=\frac{x+y+7}{z}+1=3$

$\Rightarrow \frac{x+y+z-3}{x}=\frac{x+y+z-4}{y}=\frac{x+y+z+7}{z}=3$
$\Rightarrow \frac{0,5-3}{x}=\frac{0,5-4}{y}=\frac{0,5+7}{z}=3$

$\Rightarrow x=\frac{-5}{6}; y=\frac{-7}{6}; z=\frac{5}{2}$