Question

Tìm x , y thõa mãn :

\(\left|x+\frac{8}{5}\right|+\left|2,2-2y\right|\le0\)

Answer 1

Ta luôn có : \(\left|x+\frac{8}{5}\right|\ge0\) , \(\left|2,2-2y\right|\ge0\)

Suy ra \(\left|x+\frac{8}{5}\right|+\left|2,2-2y\right|\ge0\)

mà \(\left|x+\frac{8}{5}\right|+\left|2,2-2y\right|\le0\)

Do đó : \(\left|x+\frac{8}{5}\right|+\left|2,2-2y\right|=0\)

\(\Rightarrow\begin{cases}\left|x+\frac{8}{5}\right|=0\\\left|2,2-2y\right|=0\end{cases}\) \(\Rightarrow\begin{cases}x=-\frac{8}{5}\\y=\frac{11}{10}\end{cases}\)

Answer 2

Ta có

\(\begin{cases}\left|x+\frac{8}{5}\right|\ge0\\\left|2,3-2y\right|\ge0\end{cases}\)

=> \(\left|x+\frac{8}{5}\right|+\left|2,3-2y\right|\ge0\)

=> \(x,y\in\varnothing\)

Answer 3

Vì : \(\left|x+\frac{8}{5}\right|\ge0;\left|2,2-2y\right|\ge0\)

\(\Rightarrow\left|x+\frac{8}{5}\right|+\left|2,2-2y\right|\ge0\)

Mà theo đề bài : \(\left|x+\frac{8}{5}\right|+\left|2,2-2y\right|\le0\)

\(\Rightarrow\left|x+\frac{8}{5}\right|+\left|2,2-2y\right|=0\)

\(\Rightarrow\begin{cases}\left|x+\frac{8}{5}\right|=0\\\left|2,2-2y\right|=0\end{cases}\)

\(\Rightarrow\begin{cases}x+\frac{8}{5}=0\\2,2-2y=0\end{cases}\)

\(\Rightarrow\begin{cases}x=-\frac{8}{5}\\2y=2,2\end{cases}\)

\(\Rightarrow\begin{cases}x=\frac{-8}{5}\\y=1,1\end{cases}\)

\(\Rightarrow\begin{cases}x=\frac{-8}{5}\\y=\frac{11}{10}\end{cases}\)