Question

[Đặng Vũ Đức Anh]

x+y = 1/2 , y+z= 1/3 , z+x= 1/6 Tìm x,y,z thỏa mãn :

Answer 1

[乡☪ɦαทɦ💥☪ɦųα✔]

Ta có :

\(x+y=\frac{1}{2};y+z=\frac{1}{3};z+x=\frac{1}{6}\)

\(\Rightarrow\left(x+y\right)+\left(y+z\right)+\left(z+x\right)=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\)

\(\Rightarrow2x+2y+2z=\frac{3}{6}+\frac{2}{6}+\frac{1}{6}\)

\(\Rightarrow2\left(x+y+z\right)=1\)

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

\(\Rightarrow\hept{\begin{cases}\left(x+y+z\right)-\left(x+y\right)=\frac{1}{2}-\frac{1}{2}\Rightarrow z=0\\\left(x+y+z\right)-\left(y+z\right)=\frac{1}{2}-\frac{1}{3}\Rightarrow x=\frac{1}{6}\\\left(x+y+z\right)-\left(z+x\right)=\frac{1}{2}-\frac{1}{6}\Rightarrow y=\frac{1}{3}\end{cases}}\)

Vậy \(x=\frac{1}{6},y=\frac{1}{3};z=0\) .

Answer 2

[FL.Han_]

\(x+y=\frac{1}{2};y+z=\frac{1}{3};z+x=\frac{1}{6}\)

Ta có:\(\left(x+y\right)+\left(y+z\right)+\left(z+x\right)=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\)

\(\Leftrightarrow2\left(x+y+z\right)=1\)

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

\(\Rightarrow\hept{\begin{cases}\left(x+y+z\right)-\left(x+y\right)=\frac{1}{2}-\frac{1}{2}=0\\\left(x+y+z\right)-\left(y+z\right)=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\\\left(x+y+z\right)-\left(z+x\right)=\frac{1}{2}-\frac{1}{6}=\frac{1}{3}\end{cases}}\)

Vậy....