Question

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

Tính x,y,z

Answer 1

\(\text{Ta có : }\dfrac{x}{y+z-5}=\dfrac{y}{x+z+3}=\dfrac{z}{x+y+2}\left(1\right)\)
Áp dụng tính chất dãy tỉ số bằng nhau

\(\text{ Ta được: }\dfrac{x}{y+z-5}=\dfrac{y}{x+z+3}=\dfrac{z}{x+y+2}\\ =\dfrac{x+y+z}{\left(y+z-5\right)+\left(x+z+3\right)+\left(x+y+2\right)}\\ \\=\dfrac{x+y+z}{2x+2y+2z}\\ \\=\dfrac{x+y+z}{2\left(x+y+z\right)}=\dfrac{1}{2}\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\) suy ra :

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

\(\text{Ta lại có : }\left\{{}\begin{matrix}\dfrac{x}{y+z-5}=\dfrac{1}{2}\\\dfrac{y}{x+z+3}=\dfrac{1}{2}\\\dfrac{z}{x+y+2}=\dfrac{1}{2}\end{matrix}\right.\\ \text{Kết hợp với }\left(\text{*}\right)\text{suy ra : }\left\{{}\begin{matrix}\dfrac{x}{1-x-5}=\dfrac{1}{2}\\\dfrac{y}{1-y+3}=\dfrac{1}{2}\\\dfrac{z}{1-z+2}=\dfrac{1}{2}\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{x}{-4-x}=\dfrac{1}{2}\\\dfrac{y}{4-y}=\dfrac{1}{2}\\\dfrac{z}{3-z}=\dfrac{1}{2}\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}2x=-4-x\\2y=4-y\\2z=3-z\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}3x=-4\\3y=4\\3z=3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{4}{3}\\y=\dfrac{4}{3}\\x=1\end{matrix}\right.\)

Vậy \(x=-\dfrac{4}{3};y=\dfrac{4}{3};z=1\)

\(\)

Answer 2

Áp dụng tích chất của dãy tỉ số bằng nhau, ta có

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

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

+)\(x+y+z=\dfrac{1}{2}=>x+y=\dfrac{1}{2}-z=>\dfrac{1}{2}-z+2=2z=>3z=\dfrac{5}{2}=>z=\dfrac{5}{6}\)

\(=>x+y+z=\dfrac{1}{2}+\dfrac{5}{6}+y=\dfrac{1}{2}=>\dfrac{4}{3}+y=\dfrac{1}{2}=>y=\dfrac{-5}{6}\)

Vậy \(x=\dfrac{1}{2}\\ y=\dfrac{-5}{6}\\ z=\dfrac{5}{6}\)

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