Question

[Nguyen Hai Linh]

I x-2 I +I 1.5-y I + I 3-zI=0

Answer 1

[Nguyễn Hưng Phát]

Ta có:\(\hept{\begin{cases}\left|x-2\right|\ge0\\\left|1,5-y\right|\ge0\\\left|3-z\right|\ge0\end{cases}\Rightarrow\left|x-2\right|+\left|1,5-y\right|+\left|3-z\right|\ge0}\)

Để \(\left|x-2\right|+\left|1,5-y\right|+\left|3-z\right|=0\) thì \(\hept{\begin{cases}\left|x-2\right|=0\\\left|1,5-y\right|=0\\\left|3-z\right|=0\end{cases}\Rightarrow\hept{\begin{cases}x=2\\y=1,5\\z=3\end{cases}}}\)

Answer 2

[Nguyễn Anh Quân]

Vì |x-2| ; |1,5-y| ; |3-z| đều >= 0 nên VT >= 0 

=> VT= 0 <=> x-2=0;1,5-y=0;3-z=0

<=> x=2;y=1,5;z=3