Question

Tìm x,y,z, biết:

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

Answer 1

sai đề nè: chỗ thứ 2 là x+z+1

Áp dụng tính chất dãy tỉ số bằng nhau:

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

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

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

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

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

Ta có:\(\dfrac{x}{y+z+1}=\dfrac{x}{\dfrac{1}{2}-x+1}=\dfrac{x}{\dfrac{3}{2}-x}=\dfrac{1}{2}\)<=>2x=\(\dfrac{3}{2}-x\)<=>x=\(\dfrac{1}{2}\)

\(\dfrac{y}{x+z+1}=\dfrac{y}{\dfrac{1}{2}-y+1}=\dfrac{y}{\dfrac{3}{2}-y}=\dfrac{1}{2}\)<=>2y=\(\dfrac{3}{2}-y\)<=>y=\(\dfrac{1}{2}\)

x=\(\dfrac{1}{2}\)=>y+z=0=>z=-\(\dfrac{1}{2}\)