Question

[Ankane Yuki]

Tìm x,y,z biết:

\(|1-y|+|z+2y|+|x+y+z|=0\)

Answer 1

[Arima Kousei]

Ta có : 

\(\left|1-y\right|\ge0\)

\(\left|z+2y\right|\ge0\)

\(\left|x+y+z\right|\ge0\)

\(\Rightarrow\left|1-y\right|+\left|x+2y\right|+\left|x+y+z\right|\ge0\)

Dấu \("="\)

\(\Leftrightarrow\hept{\begin{cases}\left|1-y\right|=0\\\left|z+2y\right|=0\\\left|x+y+z\right|=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}1-y=0\\z+2y=0\\x+y+z=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y=1\\z+2.1=0\\x+1+z=0\end{cases}\Leftrightarrow\hept{\begin{cases}y=1\\z+2=0\\x+1+z=0\end{cases}\Leftrightarrow}\hept{\begin{cases}y=1\\z=-2\\x+1+-2=0\end{cases}}}\)

\(\Leftrightarrow\hept{\begin{cases}y=1\\z=-2\\x=1\end{cases}}\)

Vậy ...

Answer 2

[Phùng Minh Quân]

Ta có : 

\(\hept{\begin{cases}\left|1-y\right|\ge0\\\left|z+2y\right|\ge0\\\left|x+y+z\right|\ge0\end{cases}\Rightarrow\left|1-y\right|+\left|z+2y\right|+\left|x+y+z\right|\ge0}\)

Mà \(\left|1-y\right|+\left|z+2y\right|+\left|x+y+z\right|=0\) ( giả thiết ) 

Suy ra \(\hept{\begin{cases}\left|1-y\right|=0\\\left|z+2y\right|=0\\\left|x+y+z\right|=0\end{cases}\Leftrightarrow\hept{\begin{cases}1-y=0\\z+2y=0\\x+y+z=0\end{cases}}}\)

\(\Leftrightarrow\)\(\hept{\begin{cases}y=1\\z=\left(-2\right).1\\x=\left[1+\left(-2\right).1\right]\end{cases}\Leftrightarrow\hept{\begin{cases}y=1\\z=-2\\x=-1\end{cases}}}\)

Vậy \(x=-1\)\(;\)\(y=1\) và \(z=-2\)

Chúc bạn học tốt ~