Tham khảo
`\frac{x}{y+z}=\frac{y}{z+x}=\frac{z}{x+y}=x+y+z`
Áp dụng T/C dãy tỉ số = nhau
`\frac{x}{y+z}=\frac{y}{z+x}=\frac{z}{x+y}=x+y+z=\frac{x+y+z}{y+z+z+x+x+y}=\frac{x+y+z}{2(x+y+z)}=\frac{1}{2}`
Do đó `x+y+z=\frac{1}{2}`
Xét `\frac{x}{y+z}=\frac{1}{2}⇒y+z=2x ⇒x+y+z=3x⇒\frac{1}{2}=3x⇒x=\frac{1}{2}:3=\frac{1}{6}`
Xét `\frac{y}{z+x}=\frac{1}{2}⇒z+x=2y⇒z+y+x=3y⇒\frac{1}{2}=3y⇒y=\frac{1}{2}:3=\frac{1}{6}`
Xét `\frac{z}{x+y}=\frac{1}{2}⇒x+y=2z⇒z+y+x=3z⇒\frac{1}{2}=3z⇒z=\frac{1}{2}:3=\frac{1}{6}`
Vậy `x=y=z=\frac{1}{6}`
