Question

Tìm x,y,z biết

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

Answer 1

Có:

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

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

\(\Rightarrow2x=x+y-2\)

\(\Rightarrow2z+2=x+y\) ( quy tắc chuyển vế )

Lại có:

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

\(\Leftrightarrow2z+2+z=\dfrac{1}{2}\)

\(\Leftrightarrow3z+2=\dfrac{1}{2}\)

\(\Leftrightarrow3z=\dfrac{1}{2}-2=\dfrac{-3}{2}\)

\(\Leftrightarrow z=\dfrac{-1}{2}\)

Vậy \(\left\{x,y,z\right\}=\left\{\dfrac{1}{2},\dfrac{1}{2},\dfrac{-1}{2}\right\}\)

Chúc bạn học tốt!