Question

B1 ,tìm x, y,z biết:

A , x/y+z+1 = y/x+z+1 = z/x+y+2 = x+y+z

Answer 1

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

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

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

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{y+z+1}=\dfrac{1}{2}\\\dfrac{y}{x+z+1}=\dfrac{1}{2}\\\dfrac{z}{x+y-2}=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y+z=2x\\2y=x+z\\2z=x+y-2\end{matrix}\right.\)

Cộng cả 3 vế:

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

Ko tồn tại 3 số x;y;z