Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) Ta có: \(\left|x+\dfrac{19}{5}\right|\ge0\forall x\in Q\)
\(\left|y+\dfrac{2017}{2018}\right|\ge0\forall y\in Q\)
\(\left|z-2019\right|\ge0\forall x\in Q\)
\(\Rightarrow\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{2017}{2018}\right|+\left|z-2019\right|\ge0\forall x,y,z\in Q\)
Dấu \("="\) xảy ra khi \(\left\{{}\begin{matrix}\left|x+\dfrac{19}{5}\right|=0\\\left|y+\dfrac{2017}{2018}\right|=0\\\left|z-2019\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-19}{5}\\y=\dfrac{-2017}{2018}\\z=2019\end{matrix}\right.\).
b) Lại có:
\(\left|x-\dfrac{9}{5}\right|\ge0\forall x\in Q\)
\(\left|y+\dfrac{3}{4}\right|\ge0\forall y\in Q\)
\(\left|z+\dfrac{7}{2}\right|\ge0\forall z\in Q\)
\(\Rightarrow\left|x-\dfrac{9}{5}\right|+\left|y+\dfrac{3}{4}\right|+\left|z+\dfrac{7}{2}\right|\ge0\forall x,y,zQ\)
Mà theo đề bài:
\(\left|x-\dfrac{9}{5}\right|+\left|y+\dfrac{3}{4}\right|+\left|z+\dfrac{7}{2}\right|\le0\forall\)
\(\Rightarrow\left|x-\dfrac{9}{5}\right|+\left|y+\dfrac{3}{4}\right|+\left|z+\dfrac{7}{2}\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x-\dfrac{9}{5}\right|=0\\\left|y+\dfrac{3}{4}\right|=0\\\left|z+\dfrac{7}{2}\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{9}{5}\\y=\dfrac{-3}{4}\\z=\dfrac{-7}{2}\end{matrix}\right.\)
Vậy .....
a) \(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{2017}{2018}\right|+\left|z-2019\right|=0\)
Ta có: \(\left|x+\dfrac{19}{5}\right|\ge0;\left|y+\dfrac{2017}{2018}\right|\ge0;\left|z-2019\right|\ge0\)
Để \(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{2017}{2018}\right|+\left|z-2019\right|=0\) thì:
\(\left\{{}\begin{matrix}\left|x+\dfrac{19}{5}\right|=0\\\left|y+\dfrac{2017}{2018}\right|=0\\\left|z-2019\right|=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-19}{5}\\y=\dfrac{-2017}{2018}\\z=2019\end{matrix}\right.\)
Vậy............................
b) Ta có: \(\left|x-\dfrac{9}{5}\right|\ge0;\left|y+\dfrac{3}{4}\right|\ge0;\left|z+\dfrac{7}{2}\right|\ge0\)
Mà \(\left|x-\dfrac{9}{5}\right|+\left|y+\dfrac{3}{4}\right|+\left|z+\dfrac{7}{2}\right|\le0\) thì:
\(\left|x-\dfrac{9}{5}\right|=\left|y+\dfrac{3}{4}\right|=\left|z+\dfrac{7}{2}\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{9}{5}\\y=\dfrac{-3}{4}\\z=\dfrac{-7}{2}\end{matrix}\right.\)
Vậy............................
Bài 2:
a: \(f\left(-x\right)=-x+\left|-x\right|=-x+\left|x\right|< >f\left(x\right)\)
Vậy: Hàm số không chẵn cũng không lẻ
b: \(f\left(-x\right)=-x-\left|-x\right|=-x-\left|x\right|< >f\left(x\right)\)
Vậy: Hàm số không chẵn cũng không lẻ