Question

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

Answer 1

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

Mà \(\left|x+\frac{8}{5}\right|+\left|2,2-2y\right|\le0=>\left|x+\frac{8}{5}\right|+\left|2,2-2y\right|=0\)

\(=>\hept{\begin{cases}\left|x+\frac{8}{5}\right|=0\\\left|2,2-2y\right|=0\end{cases}=>\hept{\begin{cases}x+\frac{8}{5}=0\\2,2-2y=0\end{cases}=>\hept{\begin{cases}x=-\frac{8}{5}\\2y=2,2\end{cases}=>\hept{\begin{cases}x=-1,6\\y=1,1\end{cases}}}}}\)

Vậy x = -1,6; y = 1,1

Ủng hộ mk nha ^_-