Question

Tìm x,y,z biết:\(\dfrac{x}{y+z-5}=\dfrac{y}{x+z+3}=\dfrac{z}{x+y+2}=\dfrac{1}{2}\)

Answer 1

Đề có sai không?

\(\dfrac{x}{y+z-5}=\dfrac{y}{x+z+3}=\dfrac{z}{x+y+2}=x+y+z=\dfrac{1}{2}\)

\(\Rightarrow\left\{{}\begin{matrix}y+z-5=2x\\x+z+3=2y\\x+y+2=2z\\x+y+z=\dfrac{1}{2}\end{matrix}\right.\)

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

Hay: \(\left\{{}\begin{matrix}\dfrac{1}{2}-x-5=2x\\\dfrac{1}{2}-y+3=2y\\\dfrac{1}{2}-z+2=2z\end{matrix}\right.\)

Rồi tự tính nhé :v